## 几个恒等式证明

1.证明:

$\sum_{j=1}^{n-1}{\frac{1}{1-\exp \left\{ \frac{2\pi ij}{n} \right\}}}=\frac{n-1}{2}.$

2.设$N$为自然数, $\{x\}$表示$x$的小数部分.证明$\sum_{n=1}^N{\left( \left\{ x+\frac{n}{N} \right\} -\frac{1}{2} \right)}=\left\{ Nx \right\} -\frac{1}{2}.$

$\int_0^1{\left( \left\{ ax \right\} -\frac{1}{2} \right) \left( \left\{ bx \right\} -\frac{1}{2} \right) dx}=\frac{1}{12ab}.$

3.求极限
$\lim_{m\rightarrow \infty}\lim_{n\rightarrow \infty}\int_0^1{\int_0^1{\cdots \int_0^1{\sum_{i=0}^m{\exp \left\{ -\frac{n}{\sum_{k=1}^n{x_{k}^{m-1}}} \right\} \frac{\prod_{j=1}^i{\sum_{k=1}^n{x_{k}^{j-1}}}}{\left( \sum_{k=1}^n{x_{k}^{m-1}} \right) ^i}}dx_1dx_2\cdots dx_n}}}.$
4.证明
\begin{align*}f\left( x \right) &=\frac{1}{a}+\frac{x}{a\left( a+d \right)}+\cdots +\frac{x^n}{a\left( a+d \right) \cdots \left( a+nd \right)}+\cdots\\&=\frac{e^{x/d}}{dx^{a/d}}\int_0^x{e^{-t/d}t^{a/d-1}dt}.\end{align*}

5.证明Ramanujan的恒等式
$\int_0^{\infty}{e^{-3\pi x^2}\frac{\sin\text{h}\pi x}{\sin\text{h}3\pi x}dx}=\frac{1}{\sqrt{3}e^{2\pi /3}}\sum_{n=0}^{\infty}{\frac{e^{-2n\left( n+1 \right) \pi}}{\left( 1+e^{-\pi} \right) ^2\left( 1+e^{-3\pi} \right) ^2\cdots \left( 1+e^{-\left( 2n+1 \right) \pi} \right) ^2}}.$

6.Compute
$$\sum_{n=1}^{\infty}{\left( \frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}} \right)}.$$
By the software Mathematica, I find
$$\sum_{n=1}^{\infty}{\left( \frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}} \right)}=-\frac{2}{3}-\frac{\zeta \left( 1/2 \right)}{\sqrt{2\pi}}.$$

Well, $-\frac{1}{\sqrt{2\pi}}\zeta\left(\tfrac{1}{2}\right)$ is the $\zeta$-regularization of the divergent series $\sum_{n\geq 1}\frac{1}{\sqrt{2\pi n}}$, hence the problem boils down to finding the $\zeta$-regularization of the divergent series $\sum_{n\geq 1}\frac{n^n}{n!e^n}$. As pointed out in the comments,

$$W(x) = \sum_{n\geq 1}\frac{n^{n-1}(-1)^{n-1}}{n!}x^n$$
holds for any $x\in\left(-\frac{1}{e},\frac{1}{e}\right)$ by Lagrange inversion theorem, hence
$$ze^{-z} W'(-ze^{-z})=\sum_{n\geq 1}\frac{n^n}{n!e^{nz}}z^{n}=\frac{z}{1-z} =\sum_{n\geq 1}z^n\tag{1}$$
holds for any $z\in(-W(e^{-1}),1)$. Pretty strange identity, I can give you that.<br>
Similarly, over the same interval
$$-W(-z e^{-z})=\sum_{n\geq 1}\frac{n^{n-1}}{n!e^{nz}}z^n = z \tag{2}$$
$$1=\sum_{n\geq 1}\frac{n^{n}}{n!e^{nz}}z^{n-1}-\sum_{n\geq 1}\frac{n^{n}}{n!e^{nz}}z^{n}=\frac{1}{1-z}-\frac{z}{1-z}.\tag{3}$$
Since $\zeta(0)=-\frac{1}{2}$, it should not be difficult to prove from $(1)$ and $(2)$ that the $\zeta$-regularization of $\sum_{n\geq 1}\frac{n^n}{n!e^n}$ equals $-\frac{2}{3}$ as wanted, for instance by computing $\sum_{n\geq 1}\frac{n^{n-1-k}}{n!e^n}$ for any $k\in\mathbb{N}$:
$$\sum_{n\geq 1}\frac{n^{n-2}}{n!e^n}=\int_{-1/e}^{1}\frac{W(x)}{x}\,dx = \frac{1}{2},\qquad \sum_{n\geq 1}\frac{n^{n-3}}{n!e^n}=-\int_{-1/e}^{1}\frac{W(x)}{x}(1+\log(-x))\,dx=\frac{5}{12}$$
$$\sum_{n\geq 1}\frac{n^{n-4}}{n!e^n}=\frac{7}{18},\qquad \sum_{n\geq 1}\frac{n^{n-4}}{n!e^n}=\frac{1631}{4320},$$
$$\sum_{n\geq 1}\frac{n^{n-1-k}}{n!e^n}= \frac{1}{\Gamma(k)}\int_{0}^{1}(1-x)(x-1-\log x)^{k-1}\,dx.\tag{4}$$
Indeed the substitution $x=e^{-s}$ in $(4)$ and the integral representation for the $\zeta$ function complete the proof.
Taking $$F\left(x\right)=\sum_{n\geq1}\frac{n^{n-1}}{n!e^{n}}x^{n}-\frac{1}{\sqrt{2\pi}}\sum_{n\geq1}\frac{x^{n}}{n^{3/2}}=-W\left(-\frac{x}{e}\right)-\frac{\mathrm{Li}_{3/2}\left(x\right)}{\sqrt{2\pi}},\,\left|x\right|<1$$ where $W\left(x\right)$ is the [Lambert $W$ function][1] and $\mathrm{Li}_{3/2}\left(x\right)$ is the [Polylogarithm function][2], we obtain, differentiating both sides,that $$\sum_{n\geq1}\left(\frac{n^{n}}{n!e^{n}}-\frac{1}{\sqrt{2\pi n}}\right)x^{n-1}=-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}-\frac{\mathrm{Li}_{1/2}\left(x\right)}{x\sqrt{2\pi}}$$ so $$\sum_{n\geq1}\left(\frac{n^{n}}{n!e^{n}}-\frac{1}{\sqrt{2\pi n}}\right)=\lim_{x\rightarrow1^{-}}\left(-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}-\frac{\mathrm{Li}_{1/2}\left(x\right)}{x\sqrt{2\pi}}\right).$$ Now, [we know][3] that $$\mathrm{Li}_{v}\left(z\right)=\left(\Gamma\left(1-v\right)\left(1-z\right)^{v-1}+\zeta\left(v\right)\right)\left(1+O\left(\left|1-z\right|\right)\right),v\neq1,\,z\rightarrow1$$  and now we claim $$-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}\sim\frac{1}{\sqrt{2\left(1-x\right)}}-\frac{2}{3}$$ as $x\rightarrow1^{-}$. This is true because, [since][5] $$W\left(z\right)\sim-1+\sqrt{2ze+2}-\frac{2}{3}e\left(z+\frac{1}{e}\right)$$ as $z\rightarrow-1/e$, we have $$-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}\sim\frac{1-\sqrt{2\left(1-x\right)}+\frac{2}{3}\left(1-x\right)}{x\sqrt{2\left(1-x\right)}-\frac{2}{3}\left(1-x\right)x}$$ $$=\frac{1}{x}\left(-1+\frac{1}{\sqrt{2\left(1-x\right)}}\left(\frac{1}{1-\sqrt{2-2x}/3}\right)\right)=\frac{1}{x}\left(-1+\frac{1}{\sqrt{2\left(1-x\right)}}\sum_{k\geq0}\left(\frac{\sqrt{2-2x}}{3}\right)^{k}\right)$$ $$=\frac{1}{x}\left(-\frac{2}{3}+\frac{1}{\sqrt{2\left(1-x\right)}}+O\left(\sqrt{1-x}\right)\right)$$ then the claim.

[1]:https://en.wikipedia.org/wiki/Lambert_W_function
[2]:https://en.wikipedia.org/wiki/Polylogarithm
[3]:http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/06/01/02/01/01/
[4]:http://mathworld.wolfram.com/StirlingsSeries.html
[5]:http://functions.wolfram.com/ElementaryFunctions/ProductLog/06/01/02/
This is a general answer to the followup question by Jack D'Aurizio.

**Proposition**
>Let $n\in\mathbb{N}$. We have the asymptotic expansion
$$n!\sim \sqrt{2\pi n} \frac{n^n}{e^n} \left[ 1+ \frac1{12n} +\frac1{288n^2}-\frac{139}{51840n^3}-\frac{157}{2488320n^4}+\cdots \right]$$

Note that this is not a convergent series, but an asymptotic expansion. The error in the truncated series is asymptotically equal to the first omitted term. Regard the series on the right as an element of the ring of power series over rational numbers $\mathbb{Q}[[T]]$.
$$S(T)=1+ \frac1{12}T+\frac1{288}T^2-\frac{139}{51840}T^3-\frac{157}{2488320}T^4 + \cdots.$$
Consider the multiplicative inverse of $S(T)$ in $\mathbb{Q}[[T]]$.
$$S^{-1}(T)=1-\frac1{12}T+ g_2 T^2 + g_3 T^3 + g_4 T^4 + \cdots.$$
Let $Y_s(T)=\sum_{n=0}^{\infty} h_n(s) T^n \in\mathbb{Q}[s][[T]]$ be defined by
$$\left(\frac12 T^2\right)^{s-1}\sum_{n=0}^{\infty} h_n(s) T^n = \left[ \frac12 T^2 + \frac13 T^3 + \frac14 T^4+\cdots \right]^{s-1}.$$
Then we have

**Theorem**
>$$\sum_{n=1}^{\infty} n^p\left[ \frac{n^n}{n!e^n}- \frac1{\sqrt{2\pi n}} \sum_{k=0}^p \frac{g_k}{n^k}\right]=(-2)^p p!h_{2p+1}(-p) - \frac1{\sqrt{2\pi}}\sum_{k=0}^p g_k \zeta\left(k+\frac12-p\right).$$

With $p=0$, it is the original series
$$\sum_{n=1}^{\infty} \left[\frac{n^n}{n!e^n} - \frac1{\sqrt{2\pi n}}\right]=-\frac23 - \frac{\zeta\left(\frac12\right)}{\sqrt{2\pi}}$$

With $p=1$, it gives the value of
$$\sum_{n=1}^{\infty} \left[ \frac{n^{n+1}}{n!e^n} - \sqrt{\frac{n}{2\pi}} + \frac{1}{12\sqrt{2\pi n}}\right] = -\frac 4{135} - \frac{\zeta\left(-\frac12\right)}{\sqrt{2\pi}} + \frac{\zeta\left(\frac12\right)}{12\sqrt{2\pi}}.$$

7.Prove that $$\sum_{m\leqslant x}\sum_{n\leqslant x}\Big\{\frac{x}{m+n}\Big\}=\Big(2\log2-\frac{\pi^2}{12}\Big)x^2+O(x\log x),$$

where $\{x\}$ is the fractional part of the real number $x$.

I know
$$\sum_{n\leqslant x} \Big\{\frac{x}{n} \Big\}=(1-\gamma)x+O\big(x^{1/2}\big),$$

where $\gamma$ is Euler's constant. But I don't know whether it is useful. Can you help me?

Write $\{x\} = x - \lfloor x\rfloor$,s so that
$\sum_{m \leq x} \sum_{n \leq x} \left\{ \frac{x}{m + n}\right\} = \sum_{m \leq x} \sum_{n \leq x} \frac{x}{m + n} - \sum_{m \leq x} \sum_{n \leq x} \left\lfloor \frac{x}{m + n}\right\rfloor.$
Then note that
$\sum_{m \leq x} \sum_{n \leq x} \left\lfloor \frac{x}{m + n}\right\rfloor = \sum_{m \leq x} \sum_{n \leq x} \sum_{\ell \leq \frac{x}{m + n}} 1 = \sum_{\ell \leq \frac{x}{2}} \sum_{n \leq \frac{x}{\ell} - 1} \sum_{m \leq \frac{x}{\ell} - n} 1.$
The sum over $m$ is $\lfloor x/\ell\rfloor - n$. The ensuing sum over $n$ is
$\left\lfloor \frac{x}{\ell}\right\rfloor \left(\left\lfloor \frac{x}{\ell}\right\rfloor - 1\right) - \sum_{n \leq \frac{x}{\ell} - 1} n.$
Via partial summation,
$\sum_{n \leq \frac{x}{\ell} - 1} n = \left(\frac{x}{\ell} - 1\right) \left(\left\lfloor \frac{x}{\ell} \right\rfloor - 1\right) - \int_{1}^{\frac{x}{\ell} - 1} \lfloor t\rfloor \\, dt,$
and this integral is equal to $\frac{1}{2} \left(\frac{x}{\ell} - 1\right)^2 + O(\frac{x}{\ell})$. So the sum over $n$ and $m$ simplifies to
$\frac{x^2}{2 \ell^2} + O\left(\frac{x}{\ell}\right).$
The ensuing sum over $\ell$ is
$\frac{x^2}{2} \sum_{\ell = 1}^{\infty} \frac{1}{\ell^2} - \frac{x^2}{2} \sum_{\ell > \frac{x}{2}} \frac{1}{\ell^2} + O(x \log x).$
The first sum over $\ell$ is $\zeta(2) = \pi^2/6$. The second is $O(1/x)$. So this simplifies to
$\frac{\pi^2 x^2}{12} + O(x \log x).$

Now we deal with
$\sum_{m \leq x} \sum_{n \leq x} \frac{x}{m + n} = x \sum_{m \leq x} \sum_{n \leq x} \frac{1}{m + n}.$
We deal with the sum over $n$ via partial summation: it is equal to
$\frac{\lfloor x\rfloor}{m + x} + \int_{1}^{x} \frac{\lfloor t \rfloor}{(m + t)^2} \\, dt = \log \frac{m + x}{m + 1} + O\left(\frac{1}{m}\right),$
where we have used the fact that $\lfloor x \rfloor = x - \{x\} = x + O(1)$, the fact that the antiderivative of $t/(m + t)^2$ is $m/(m + t) + \log(m + t)$, and the fact that the antiderivative of $1/(m + t)^2$ is $-1/(m + t)$.

So it remains to evaluate
$\sum_{m \leq x} \log \frac{m + x}{m + 1} = \lfloor x\rfloor \log \frac{2x}{x + 1} + (x + 1) \int_{1}^{x} \frac{\lfloor t\rfloor}{(t + x)(t + 1)} \\, dt.$
The antiderivative of $\frac{t}{(t + x)(t + 1)}$ is $\frac{x}{x - 1} \log \frac{t + x}{t + 1}$, and so after some simplification, we arrive at $(2\log 2) x + O(\log x)$.

## 1-1+1-1+...＝1/2？

Suppose $f(x)>0$, $f''(x)\le 0$, and $f(+\infty)=+\infty$ on $[0,+\infty)$, prove the following conclusion
$\lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\frac{1}{2}.$

\begin{proof}
Since $f(x)>0$, $f''(x)\le 0$, and $f(+\infty)=+\infty$, it is not hard to see $f'(x)>0$, then
$\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\sum\limits_{n=0}^{\infty} \left( \frac{1}{f^s(2n)}-\frac{1}{f^s(2n+1)} \right)=-\sum\limits_{n=0}^{\infty} g_s(\xi_n),$
where $g_s(x)=(\frac{1}{f^s(x)})'=-\frac{sf'(x)}{f^{s+1}(x)}$ and $\xi_n \in (2n,2n+1)$. Since
$g_s'(x)=s\cdot \frac{(s+1)f'^2(x)-f''(x)f(x)}{f^{s+2}(x)}>0,$
we can get
$-\sum\limits_{n=0}^{\infty} g_s(\xi_n)\le -\sum\limits_{n=0}^{\infty} g_s(2n)\le -g_s(0)-\frac{1}{2}\sum\limits_{n=1}^{\infty} \int_{2n-2}^{2n} g_s(x)dx$
$=-g_s(0)-\frac{1}{2}\int_{0}^{+\infty} g_s(x)dx=-g_s(0)+\frac{1}{2f^s(0)}\to \frac{1}{2}$
as $s\to 0_+$. Similarly, we have
$-\sum\limits_{n=0}^{\infty} g_s(\xi_n)\ge -\sum\limits_{n=0}^{\infty} g_s(2n+1)\ge -\frac{1}{2}\sum\limits_{n=0}^{\infty} \int_{2n+1}^{2n+3} g_s(x)dx$
$=-\frac{1}{2}\int_{1}^{+\infty} g_s(x)dx=\frac{1}{2f^s(1)}\to \frac{1}{2}$
as $s\to 0_+$.
\end{proof}

Prove that $\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sn^2}=\frac{1}{2}.$

\begin{proof}
Let $f_s(x)=e^{-sx^2}$, by Euler-Maclaurin formula, we have
$\sum\limits_{n=0}^{N} f_s(n)=\int_{0}^{N} f_s(x)dx+\frac{f_s(0)+f_s(N)}{2}+\frac{f_s'(0)+f_s'(N)}{12}-\frac{1}{2}\int_{0}^{N} B_2(x)f_s''(x)dx.$
Since $f_s'(x)=-2sxe^{-sx^2}$, $f_s''(x)=2se^{-sx^2}(2sx^2-1)$, we can get
$\int_{0}^{N} f_s(x)dx\to \int_{0}^{+\infty} f_s(x)dx=\int_{0}^{+\infty} e^{-sx^2}dx=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}}$
$f_s(0)+f_s(N)=1+e^{-sN^2}\to 1$
$f_s'(0)+f_s'(N)=-2sNe^{-sN^2}\to 0$
$\int_{0}^{N} B_2(x)f_s''(x)dx\to \int_{0}^{\infty} B_2(x)f_s''(x)dx=\int_{0}^{\infty} B_2(x)2se^{-sx^2}(2sx^2-1)dx$
$=2\sqrt{s}\int_{0}^{\infty} B_2(\frac{x}{\sqrt{s}})e^{-x^2}(2x^2-1)dx$
as $N\to \infty$. Therefore, we have
$\sum\limits_{n=0}^{\infty} e^{-sn^2}=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}}+\frac{1}{2}-2\sqrt{s}\int_{0}^{\infty} B_2(\frac{x}{\sqrt{s}})e^{-x^2}(2x^2-1)dx.$
Since $B_2(x)$ is bounded, we can get $\sum\limits_{n=0}^{\infty} e^{-sn^2}=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}}+\frac{1}{2}+O(\sqrt{s})$ as $s\to 0_+$, then $\sum\limits_{n=0}^{\infty} (-1)^ne^{-sn^2}=2\sum\limits_{n=0}^{\infty} e^{-4sn^2}-\sum\limits_{n=0}^{\infty} e^{-sn^2}=\frac{1}{2}+O(\sqrt{s})$.\\
Another method: Let $\vartheta(t)=\sum\limits_{n\in\mathbb{Z}}e^{-\pi n^2t}$, $t>0$ be the Jocabi theta function. Since
$\sum\limits_{n=1}^{\infty} e^{-\pi n^2t}\le \sum\limits_{n=1}^{\infty} e^{-\pi nt}=\frac{e^{-\pi t}}{1-e^{-\pi t}}=O(e^{-\pi t}),$
we can get
$\vartheta(t)=1+2\sum\limits_{n=1}^{\infty} e^{-\pi n^2t}=1+O(e^{-\pi t})$
and
$\vartheta(\frac{1}{t})=\sqrt{t}\theta(t)=\sqrt{t}+O(\sqrt{t}e^{-\pi t}).$
as $t\to \infty$. Then we have
$\sum\limits_{n=0}^{\infty} e^{-sn^2}= \frac{1}{2}+\frac{1}{2}\vartheta(\frac{s}{\pi})=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}}+\frac{1}{2}+O(s^{-\frac{1}{2}}e^{-\frac{1}{s}})$
as $s\to 0_+$.
\end{proof}

Suppose $p(x)=x^m+a_{m-1}x^{m-1}+...+a_1x+a_0$ is a monic polynomial with degree $m\ge 1$, prove that
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sp(n)}=\frac{1}{2}.$

Prove that $\lim\limits_{s\to 0+}\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n^s}=-\frac{1}{2}.$

\begin{proof}
Since
$\Gamma(s)=\int_0^{\infty}x^{s-1}e^{-x}dx=\int_0^{\infty}(at)^{s-1}e^{-at}dat=a^s\int_0^{\infty}t^{s-1}e^{-at}dt,$
we have
$a^{-s}=\frac{1}{\Gamma(s)}\int_0^{\infty}t^{s-1}e^{-at}dt.$
Then
$\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n^s}=\frac{1}{\Gamma(s)}\int_0^{\infty}t^{s-1}\sum\limits_{n=1}^{\infty}(-1)^ne^{-nt}dt=\frac{1}{\Gamma(s)}\int_0^{\infty}t^{s-1}\frac{-e^{-t}}{1+e^{-t}}dt$
$=-\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{t^{s-1}}{1+e^t}dt=-\frac{\int_0^{\infty}\frac{t^{s-1}}{1+e^t}dt}{\int_0^{\infty}t^{s-1}e^{-t}dt}\to -\frac{1}{2}$
as $s\to 0_+$.
\end{proof}

Let $S(s)=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{1+n^2s}}$, prove that $\lim\limits_{s\to 0+}S(s)=\frac{1}{2}$.

\begin{proof}
It is generally known that $\sin z=z\mathop{\Pi}\limits_{n=1}^{\infty}(1-\frac{z^2}{n^2\pi^2})$, or
$\sin \pi z=\pi z\mathop{\Pi}\limits_{n=1}^{\infty}(1-\frac{z^2}{n^2}).$
Take logarithmic derivative, we have
$\pi\cot \pi z=\frac{1}{z}+\sum\limits_{n=1}^{\infty}\frac{-2z}{n^2-z^2}=\frac{1}{z}+\sum\limits_{n=1}^{\infty}(\frac{1}{z+n}+\frac{1}{z-n})=\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}(\frac{1}{z+n}+\frac{1}{z-n}).$
Since
$\frac{1}{\sin \pi z}-\cot \pi z=\frac{1-\cos \pi z}{\sin \pi z}=\tan \frac{\pi z}{2}$
and
$\pi \tan \frac{\pi z}{2}=\pi\cot (\pi \frac{1-z}{2})=\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}(\frac{1}{\frac{1-z}{2}+n}+\frac{1}{\frac{1-z}{2}-n})$
$=-\sum\limits_{n\in \mathbb{Z}} \left(\frac{1}{z-(2n+1)}+\frac{1}{z+(2n-1)}\right)=-\sum\limits_{n\in \mathbb{Z}} \left(\frac{1}{z-(2n+1)}+\frac{1}{z+(2n+1)}\right)$
we have
$\frac{\pi}{\sin \pi z}=\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}(-1)^n(\frac{1}{z+n}+\frac{1}{z-n})=z\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{z^2-n^2},$
then
$\frac{\pi}{\sinh \pi z}=\frac{i\pi}{\sin i\pi z}=z\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{z^2+n^2}.$
It is easy to show
$\int_{0}^{\infty}\frac{\pi dt}{\sinh(\pi y \cosh t)}=\int_{0}^{\infty} y \cosh t\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{y^2\cosh^2t+n^2}dt$
$=\int_{0}^{\infty} \sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{y^2\sinh^2t+(y^2+n^2)}d(y\sinh t)$
$=\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{y^2+n^2}}\arctan\frac{y\sinh t}{\sqrt{y^2+n^2}}|_{0}^{\infty} =\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{y^2+n^2}}\cdot\frac{\pi}{2}.$
Let $s=\frac{1}{y^2}$, then we have $y\to+\infty$ as $s\to 0_+$, and
$S(s)=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{1+n^2s}}=\frac{1}{2}+\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{1+n^2s}}=\frac{1}{2}+\frac{y}{2}\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{y^2+n^2}}$
$=\frac{1}{2}+\frac{1}{\pi}\int_{0}^{\infty}\frac{\pi y dt}{\sinh(\pi y \cosh t)}.$
\\
Since
$\frac{\pi y}{\sinh(\pi y \cosh t)}\to 0$
as $y\to \infty$, and
$\frac{\pi y}{\sinh(\pi y \cosh t)}\le\frac{1}{\cosh t}\in L^1[0,\infty),$
we have
$\int_{0}^{\infty}\frac{\pi y dt}{\sinh(\pi y \cosh t)}\to 0$
as $y\to\infty$ by dominated convergence theorem.

\end{proof}

Prove that $\lim\limits_{s\to 0+}\sum\limits_{n=1}^{\infty} (-1)^n(\frac{\sin ns}{ns})^2=-\frac{1}{2}$.\\ \\ \\ \\

Suppose $a_s(x)=a(x,s)\in C([0,\infty)\times[0,1])$ and $a_s(x)\in C^1[0,\infty)$. Under the following conditions\\
$\text{(i)}a_0(x)=1,$\\
$\text{(ii)}a_s(\infty)=0 \text{ for } s>0,$\\
$\text{(iii)}\int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx \to 0 \text{ as } s\to 0_+,$\\
we have
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^na_s(n)=\frac{1}{2}.$

\begin{proof}
Since
$a_s(0)=\int_{0}^{\infty}a_s'(x)dx=\sum\limits_{n=0}^{\infty}\int_{n}^{n+1}a_s'(x)dx,$
we have
$\sum\limits_{n=0}^{\infty} (-1)^na_s(n)=\sum\limits_{n=0}^{\infty} \left( a_s(2n)-a_s(2n+1) \right)=\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1}a_s'(x)dx$
$=\frac{1}{2}a_s(0)+\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1}a_s'(x)dx-\frac{1}{2}\sum\limits_{n=0}^{\infty}\int_{n}^{n+1}a_s'(x)dx$
$=\frac{1}{2}a_s(0)+\frac{1}{2}\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1} \left( a_s'(x)-a_s'(x+1) \right) dx.$
Since $a_s(0)\to a_0(0)=1$ and
$|\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1} \left( a_s'(x)-a_s'(x+1) \right) dx|\le \sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1} |a_s'(x)-a_s'(x+1)| dx$
$\le \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx \to 0$
as $s\to 0_+$, we can get the conclusion.

\end{proof}

Suppose $a_s(x)=a(x,s)\in C([0,\infty)\times[0,1])$ and $a_s(x)\in C^1[0,\infty)$. Under the following conditions\\
$\text{(i)}a_0(x)=1,$\\
$\text{(ii)}a_s(\infty)=0 \text{ for } s>0,$\\
$(\text{iii}')a_s(x) \text{ is a convex function },$\\
we have
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^na_s(n)=\frac{1}{2}.$

\begin{proof}
Since $a_s(x)$ is a convex function, $a_s'(x)$ is increasing, we have
$\int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=\int_{0}^{\infty} a_s'(x+1)-a_s'(x)dx=a_s(0)-a_s(1)\to 1-1=0$
as $s\to 0_+$, it implys (iii).
\end{proof}

Suppose $f(x)>0$, $f''(x)\le 0$, and $f(\infty)=\infty$ on $[0,\infty)$, prove the following conclusion
$\lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\frac{1}{2}.$

\begin{proof}
If we take $a_s(x)=\frac{1}{f^s(x)}$, then (i), (ii) are trivial, and
$a_s''(x)=s\cdot \frac{(s+1)f'^2(x)-f''(x)f(x)}{f^{s+2}(x)}>0$
means $a_s(x)$ is convex.
\end{proof}

Suppose $f(x)>0$, $f''(x)\le 0$, and $f(\infty)=\infty$ on $[0,\infty)$, we have the following conclusion
$\lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} (-1)^ne^{-sf(n)}=\frac{1}{2}.$

\begin{proof}
If we take $a_s(x)=e^{-sf(x)}$, then (i), (ii) are trivial, and
$a_s''(x)=s(sf'^2(x)-f''(x))e^{-sf(x)}\ge 0$
means $a_s(x)$ is convex.
\end{proof}

Suppose $f\in C^{m+1}[0,\infty)$ with $f>0$, $f^{(m+1)}\le 0$, where $m\ge 1$. If $f(\infty)=\infty$, prove the following conclusion
$\lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} (-1)^ne^{-sf(n)}=\frac{1}{2}.$

\begin{proof}
We have solved the case $m=1$ before, and we only consider $m\ge 2$. Without loss of generality, assume $f, f',\cdots, f^{(m)}>0$. Otherwise, we can replace $m$ by a smaller number. If we take $a_s(x)=\frac{1}{f^s(x)}$, then (i), (ii) are trivial, and
$a_s''(x)=s(sf'^2(x)-f''(x))e^{-sf(x)}.$
By lemma, we have
$(\frac{f'^2}{f''})'\ge (1+\frac{1}{m-1}-\varepsilon)f'>0$
and
$\frac{f'^2}{f''}\ge (1+\frac{1}{m-1}-\varepsilon)f-C,$
then $\frac{f''}{f'^2}$ is decreasing to $0$, so the number of roots of $a_s''(x)$ is no more than $1$. If $s$ is small enough, then there is $x_s$ s.t. $a_s''(x_s)=0$, i.e. $s=\frac{f''(x_s)}{f'^2(x_s)}$, and
$\int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=\int_{0}^{x_s} a_s'(x)-a_s'(x+1)dx+\int_{x_s}^{\infty} a_s'(x+1)-a_s'(x)dx$
$=-a_s(0)+a_s(1)+2(a_s(x_s)-a_s(x_s+1)).$
Now, we only need to prove that $a_s(x_s)-a_s(x_s+1)\to 0$ as $s\to 0_+$. Since $a_s(x_s+1)-a_s(x_s)=a_s'(\xi_s)$, where $\xi_s\in(x_s, x_s+1)$, we have
$|a_s(x_s+1)-a_s(x_s)|\le sf'(\xi_s).$
Since
$f'(\xi_s)=f'(x_s)+\cdots+\frac{f^{(m)}(x_s)}{(m-1)!}(\xi_s-x_s)^{m-1}+\frac{f^{m+1}(\eta_s)}{m!}(\xi_s-x_s)^m$
$\le f'(x_s)+\cdots+\frac{f^{(m)}(x_s)}{(m-1)!}<<f'(x_s)$
and
$s=\frac{f''(x_s)}{f'^2(x_s)}<<\frac{1}{f(x_s)},$
we have $sf'(\xi_s)<<\frac{f'(x_s)}{f(x_s)}\to 0$ as $s\to 0_+$.
\end{proof}

Suppose $f(x)>0$, $f'''(x)\le 0$, and $f(\infty)=\infty$ on $[0,\infty)$, we have the following conclusion
$\lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} (-1)^ne^{-sf(n)}=\frac{1}{2}.$

\begin{proof}
If there is $x_0\ge 0$ s.t. $f''(x_0)\le 0$, then $f''(x)\le 0$ for $x\ge x_0$, we have solved this case. So we can assume $f''(x)>0$, and $f'(x)>0$. If we take $a_s(x)=e^{-sf(x)}$, then (i), (ii) are trivial, and
$a_s''(x)=s(sf'^2(x)-f''(x))e^{-sf(x)}.$
Since $f''(x)$ is decreasing, and $f'(x)$ is strictly increasing, then $\frac{f''(x)}{f'^2(x)}$ is decreasing to $0$, so the number of roots of $a_s''(x)$ is no more than $1$.\\
If $a_s''(x)>0$ for all $x\ge 0$, then $a_s(x)$ is convex.\\
If there is $x_s$ s.t. $a_s''(x_s)=0$, i.e. $s=\frac{f''(x_s)}{f'^2(x_s)}$, then
$\int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=\int_{0}^{x_s} a_s'(x)-a_s'(x+1)dx+\int_{x_s}^{\infty} a_s'(x+1)-a_s'(x)dx$
$=-a_s(0)+a_s(1)+2(a_s(x_s)-a_s(x_s+1)).$
Since $a_s(x_s+1)-a_s(x_s)=a_s'(\xi_s)$, where $\xi_s\in(x_s, x_s+1)$, we have
$|a_s(x_s+1)-a_s(x_s)|\le sf'(\xi_s).$
Since $f'(\xi_s)-f'(x_s)=f''(\eta_s)(\xi_s-x_s)\le f''(0)$, where $\eta_s \in(x_s, \xi_s)$, and
$f'(x_s)=\sqrt{\frac{f''(x_s)}{s}}\le \sqrt{\frac{f''(0)}{s}},$
we have
$|a_s(x_s+1)-a_s(x_s)|\le s(f''(0)+\sqrt{\frac{f''(0)}{s}})=sf''(0)+\sqrt{sf''(0)}\to 0$
as $s\to 0_+$.
\end{proof}

Prove that
$$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{n!^s}=\frac{1}{2}.$$

Lemma: Suppose $f\in C^{m+1}[0,\infty)$ and $f, f',\cdots, f^{(m)}>0$, $f^{(m+1)}\le 0$, where $m\ge 1$. Prove that
$$\limsup\limits_{x\to +\infty} \frac{f(x)f''(x)}{f'^2(x)}\le 1-\frac{1}{m}.$$

\begin{proof}
We can deal with this problem by induction. The case $m=1$ is trivial, so we can assume $m\ge 2$. Since
$f(x)=f(0)+f'(0)x+\frac{f''(\xi)}{2}x^2\ge f'(0)x,$
we can get $f(x)\to\infty$ as $x\to \infty$. Fixed $\varepsilon>0$, we have
$\frac{f'f'''}{f''^2}\le 1-\frac{1}{m-1}+\varepsilon$
for $x\ge x_0$ by induction. Since
$(\frac{f'^2}{f''})'=\frac{2f'f''^2-f'^2f'''}{f''^2}=(2-\frac{f'f'''}{f''^2})f'\ge (1+\frac{1}{m-1}-\varepsilon)f',$
we have
$\frac{f'^2}{f''}\ge (1+\frac{1}{m-1}-\varepsilon)f-C,$
that means
$\frac{ff''}{f'^2}\le \frac{f}{(1+\frac{1}{m-1}-\varepsilon)f-C}\to \frac{1}{1+\frac{1}{m-1}-\varepsilon}$
as $x\to\infty$. Let $\varepsilon\to 0$, we can obtain
$$\limsup\limits_{x\to +\infty} \frac{ff''}{f'^2}\le 1-\frac{1}{m}.$$
\end{proof}

Suppose $f\in C^{m+1}[0,\infty)$ with $f>0$, $f^{(m+1)}\le 0$, where $m\ge 1$. If $f(\infty)=\infty$, prove the following conclusion
$\lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\frac{1}{2}.$

\begin{proof}
Without loss of generality, assume $f, f',\cdots, f^{(m)}>0$. Otherwise, we can replace $m$ by a smaller number. If we take $a_s(x)=\frac{1}{f^s(x)}$, then (i), (ii) are trivial, and
$a_s''(x)=s\cdot \frac{(s+1)f'^2(x)-f''(x)f(x)}{f^{s+2}(x)}.$
By lemma, we can get $a_s''(x)>0$ for $x\ge x_0$, that means $a_s(x)$ is convex.
\end{proof}

Suppose $p(x)=x^m+a_{m-1}x^{m-1}+...+a_1x+a_0$ is a monic polynomial with degree $m\ge 1$, prove that
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sp(n)}=\frac{1}{2}.$

\begin{proof}
Let $a_s(x)=e^{-sp(x)}$, then (i), (ii) are trivial. We only need to show that
$\lim\limits_{s\to 0+}\sum\limits_{n=2N}^{\infty} (-1)^ne^{-sp(n)}=\frac{1}{2}$
for some positive integer $N$, or equivalent
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sp(n+2N)}=\frac{1}{2},$
so we can consider $p(x+2N)$ instead of $p(x)$. Without loss of generality, we can assume the coefficients of $p(x)$ are positive.

Since $a_s'(x)=-sp'(x)e^{-sp(x)}<<sx^{m-1}e^{-sx^m}$, we have $a_s(x)-a_s(x+1)=-a_s'(\xi)<<sx^{m-1}e^{-sx^m}$, where $\xi\in (x,x+1)$. Then
$\int_{0}^{\infty} |sp'(x)e^{-sp(x)}-sp'(x)e^{-sp(x+1)}|dx<<s^2\int_{0}^{\infty} x^{2(m-1)}e^{-sx^m}dx$
$=s^{\frac{1}{m}}\int_{0}^{\infty} x^{2(m-1)}e^{-x^m}dx \to 0$
as $s\to 0_+$. Since $p'(x)-p'(x+1)<<x^{m-2}$, we have
$\int_{0}^{\infty} |sp'(x)e^{-sp(x+1)}-sp'(x+1)e^{-sp(x+1)}|dx<<s\int_{0}^{\infty} x^{m-2}e^{-sx^m}dx$
$=s^{\frac{1}{m}}\int_{0}^{\infty} x^{m-2}e^{-x^m}dx \to 0$
as $s\to 0_+$. That means (iii).
\end{proof}

Suppose $f\in C^1[0,\infty)$, $f'\in L^1[0,\infty)$, and $f(0)=1$, $f(\infty)=0$, then we have
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}.$

\begin{proof}
Let $a_s(x)=f(sx)$, then (i), (ii) are trivial. Since $a_s'(x)=sf'(sx)$, we have
$\int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=s\int_{0}^{\infty} |f'(sx)-f'(s(x+1))|dx$
$=\int_{0}^{\infty} |f'(x)-f'(s+x)|dx=||f'-f_s'||_{L^1} \to 0$
as $s\to 0_+$.
\end{proof}

Let $f(x)=\frac{1}{\sqrt{1+x^2}}, (\frac{\sin x}{x})^2, 2(\frac{1}{x}-\frac{1}{e^x-1})$ and so on, then we have
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}.$\\

Suppose $F(x)$ is the Cantor-Lebesgue function, let $f(x)=F(1-x)$, then it is obvious that
$\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=1$
for $s=\frac{1}{3^n}$, and
$\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}$
for $s=\frac{2}{3^n}$, where $n\ge 1$.\\

Suppose $g(x), 0\le x\le 1$ is the unique continuous solution of the following equation

$g(x)=\frac{1}{4}f(2x), 0\le x\le\frac{1}{2},$\\

$g(x)=\frac{1}{4}+\frac{3}{4}f(2x-1), \frac{1}{2}\le x\le 1.$\\
or equivalent,

$g: \sum\limits_{n=1}^{\infty} \frac{a_n}{2^n}\mapsto \sum\limits_{n=1}^{\infty} \frac{a_n 3^{S_{n-1}}}{4^n}$
where $a_n=0, 1$, and $S_n=\sum\limits_{i=1}^{n} a_i$, $S_0=0$. Let $f(x)=g(1-x)$, then it is obvious that
$\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{3}{4}$
for $s=\frac{1}{2^n}$, where $n\ge 1$.\\

Suppose $f\in C[0,\infty)$ is an decreasing function with $f(0)=1$ and $f(\infty)=0$. If the limit
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\lambda$
exists, then $\lambda=\frac{1}{2}$.

Suppose $f\in C[0,\infty)$ is an increasing function with $f(0)=1$ and $f(\infty)=0$, then
$\liminf\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)\le\frac{1}{2}\le \limsup\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns).$

\begin{proof}
Let $f_1(x)=\int_{0}^{1}f(xt)dt=\frac{1}{x}\int_{0}^{x} f(t)dt$, then we can get the conclusion.
\end{proof}

Let $f(x)=\frac{\sin\sqrt{x}}{\sqrt{x}}$, then we have
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}.$

Suppose $f\in C^2[0,\infty)$, $f''\in L^1[0,\infty)$, and $f(0)=1$, $f(\infty)=0$, then we have
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}.$
\begin{proof}
Since $f''\in L^1[0,\infty)$, $f'(\infty)$ exists, then we have $f'(\infty)=0$ by $f(\infty)=0$. Let $f_s(x)=f(sx)$, by Euler-Maclaurin formula, we have
$\sum\limits_{n=0}^{N} f_s(n)=\int_{0}^{N} f_s(x)dx+\frac{f_s(0)+f_s(N)}{2}+\frac{f_s'(0)+f_s'(N)}{12}-\frac{1}{2}\int_{0}^{N} B_2(x)f_s''(x)dx.$
Since
$f_s(0)=f(0)=1$, $f_s'(x)=sf'(sx)$, $f_s''(x)=s^2f''(sx)$,
$\int_{0}^{N} f_s(x)dx=\frac{1}{s}\int_{0}^{sN}f(x)dx,$
$\int_{0}^{N} B_2(x)f_s''(x)dx=s^2\int_{0}^{N} B_2(x)f''(sx)dx=s\int_{0}^{sN} B_2(\frac{x}{s})f''(x)dx$
we have
$\sum\limits_{n=0}^{N} f_{2s}(n)=\frac{1}{2s}\int_{0}^{2sN}f(x)dx+\frac{1+f(2sN)}{2}+\frac{sf'(0)+sf'(2sN)}{6}$
$-s\int_{0}^{2sN} B_2(\frac{x}{2s})f''(x)dx$
and
$\sum\limits_{n=0}^{2N} f_{s}(n)=\frac{1}{s}\int_{0}^{2sN}f(x)dx+\frac{1+f(2sN)}{2}+\frac{sf'(0)+sf'(2sN)}{12}$
$-\frac{s}{2}\int_{0}^{2sN} B_2(\frac{x}{s})f''(x)dx$
Since $f''\in L^1[0,\infty)$, we can get $f'$ is bounded.
Thus,
$\sum\limits_{n=0}^{2N} (-1)^nf(ns)=2\sum\limits_{n=0}^{N} f(2ns)-\sum\limits_{n=0}^{2N} f(ns)=2\sum\limits_{n=0}^{N} f_{2s}(n)-\sum\limits_{n=0}^{2N} f_s(n)$
$=\frac{1+f(2sN)}{2}+\frac{sf'(0)+sf'(2sN)}{4}-2s\int_{0}^{2sN} B_2(\frac{x}{2s})f''(x)dx$
$+\frac{s}{2}\int_{0}^{2sN} B_2(\frac{x}{s})f''(x)dx$
Let $N\to \infty$, we have
$\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}+s\frac{f'(0)}{4}-2s\int_{0}^{\infty} B_2(\frac{x}{2s})f''(x)dx+\frac{s}{2}\int_{0}^{\infty} B_2(\frac{x}{s})f''(x)dx$
Since $B_2(x)$ is bounded, we can get $\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}+O(s)$.
\end{proof}

Let $f(x)=\frac{\sin x}{x}$, then we have
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}.$

\begin{proof}
Let $f_N(x)=f(x)\chi_{[0,N\pi]}$, it is not difficult to show that
$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf_N(ns)=\frac{1}{2},$
then we have
$\limsup\limits_{s\to 0+}|\sum\limits_{n=0}^{\infty} (-1)^nf(ns)-\frac{1}{2}|=\limsup\limits_{s\to 0+}|\sum\limits_{ns>N\pi}^{\infty} (-1)^nf(ns)|$
$=\limsup\limits_{s\to 0+}|\sum\limits_{ns>N\pi}^{\infty} \frac{(-1)^n\sin ns}{ns}|$
By Abel formula, take $a_n=(-1)^n\sin ns=\sin n(\pi+s)$, $b_n=\frac{1}{ns}$, we have
$|\sum\limits_{ns>N\pi}^{\infty} \frac{(-1)^n\sin ns}{ns}|\le \frac{A}{N\pi}.$
Let $N\to \infty$, we can get the conclusion.
\end{proof}

$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-se^n}=\frac{1}{2}?$

For any $a>1$, the limit
$\lim\limits_{x\to 1-}\sum\limits_{n=0}^{\infty} (-1)^nx^{a^n}$
does not exist.

(2018年中科大考研题) Suppose $a_n>0$ and
$\left| \sum\limits_{n=1}^{\infty} \frac{\sin (a_nx)}{n^2} \right|\le |\tan x|$
for $x\in (-1,1)$, prove that $a_n=o(n^2)$ as $n\to\infty$.
\begin{proof}
Since $\sum\limits_{n=1}^{\infty} \cfrac{\sin (a_nx)}{n^2}$ is uniform convergence, we have
$\int_{0}^{t}\sum\limits_{n=1}^{\infty} \frac{\sin (a_nx)}{n^2}dx=\sum\limits_{n=1}^{\infty} \frac{1}{n^2}\int_{0}^{t} \sin (a_nx)dx=\sum\limits_{n=1}^{\infty} \frac{1-\cos(a_nt)}{n^2a_n}$
and
$\int_{0}^{t} \tan x dx=-\ln\cos t,$
then we can get
$\sum\limits_{n=1}^{\infty} \frac{1-\cos(a_nt)}{n^2a_n}\le -\ln\cos t$
for $t\in (0,1)$, so that
$2\sum\limits_{n=1}^{N} \frac{1-\cos(a_nt)}{n^2a_nt^2}\le -\frac{2\ln\cos t}{t^2}$
for any $N\in\mathbb{N}$. Let $t\to 0_+$, then
$\sum\limits_{n=1}^{N} \frac{a_n}{n^2}\le 1.$
Let $N\to\infty$, it is obvious that
$\sum\limits_{n=1}^{\infty} \frac{a_n}{n^2}\le 1.$
\end{proof}

## 第一届熊赛分析与方程部分试题

\documentclass[11pt,a4paper]{article}
\RequirePackage{xeCJK}
%\usepackage{amsmath,amssymb,amsthm}
%%%修改数学字体
\usepackage{fontspec}
\usepackage[T1]{fontenc}
\usepackage{times}
\usepackage[complete,amssymbols,amsbb,eufrak,nofontinfo,
subscriptcorrection,zswash,mtpscr]{mtpro2}

%\usepackage{mathpazo}
%\renewcommand{\rmdefault}{ibh}
%\usepackage{fourier}
%\usepackage{charter}
%\usepackage{helvet}
\usepackage{amsmath,amsthm}
%%%修改数学字体
%\usepackage{CJKnumb}

\setCJKmainfont[BoldFont={方正黑体简体},ItalicFont={方正楷体简体}]{方正书宋简体}
%\setCJKsansfont[BoldFont={黑体}]{方正中等线简体}
%\setCJKmonofont{方正中等线简体}

\setCJKfamilyfont{kd}{华文行楷}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newcommand{\chaoda}{\fontsize{55pt}{\baselineskip}\selectfont}
\newcommand{\chuhao}{\fontsize{42pt}{\baselineskip}\selectfont}     % 字号设置
\newcommand{\xiaochuhao}{\fontsize{36pt}{\baselineskip}\selectfont} % 字号设置
\newcommand{\yihao}{\fontsize{28pt}{\baselineskip}\selectfont}      % 字号设置
\newcommand{\erhao}{\fontsize{21pt}{\baselineskip}\selectfont}      % 字号设置
\newcommand{\xiaoerhao}{\fontsize{18pt}{\baselineskip}\selectfont}  % 字号设置
\newcommand{\sanhao}{\fontsize{15.75pt}{\baselineskip}\selectfont}  % 字号设置
\newcommand{\xiaosanhao}{\fontsize{15pt}{\baselineskip}\selectfont} % 字号设置
\newcommand{\sihao}{\fontsize{14pt}{\baselineskip}\selectfont}      % 字号设置
\newcommand{\xiaosihao}{\fontsize{12pt}{14pt}\selectfont}           % 字号设置
\newcommand{\wuhao}{\fontsize{10.5pt}{12.6pt}\selectfont}           % 字号设置
\newcommand{\xiaowuhao}{\fontsize{9pt}{11pt}{\baselineskip}\selectfont}   % 字号设置
\newcommand{\liuhao}{\fontsize{7.875pt}{\baselineskip}\selectfont}  % 字号设置
\newcommand{\qihao}{\fontsize{5.25pt}{\baselineskip}\selectfont}    % 字号设置

\usepackage{ifthen}
\usepackage{eso-pic}
%\usepackage{esvect}
\usepackage{graphicx}
\usepackage{color}
%\usepackage[table,dvipsnames,svgnames]{xcolor}
\colorlet{backg}{green!20}

\usepackage{tikz}
\usepackage{makeidx}
\usepackage{enumitem}
\usepackage[left=2.6cm,right=2.6cm,top=2.54cm,bottom=2.54cm]{geometry}
\usepackage{mathtools}
\usepackage{mathrsfs}

\usepackage{multirow,booktabs}
\usepackage{makecell}
%\title{中国科学技术大学\\
%2015年硕士学位研究生入学考试试题}
%\author{(线性代数与解析几何)}
%\date{}

\newenvironment{newproof}{\par\CJKfamily{xiao}\noindent{\makebox[0pt][r]{\;\;}\textbf{证明:}}\color{black!90}\small}{\hfill$\Box$\quad\par}

\newenvironment{solution}{\par\CJKfamily{xiao}\noindent{\makebox[0pt][r]{\;\;}\textbf{解:}}\small}{\hfill$\Box$\quad\par}

\newcommand{\Ker}{\mathrm{Ker}\,}

\usepackage{fancyhdr}
\usepackage{fancybox}
\usepackage{lastpage}%此宏包是获取总页数用的。
\pagestyle{fancy}
\renewcommand{\footrulewidth}{1pt}%设置页脚线

\fancyhf{}%清除所有页眉页脚

\lfoot{科目名称：分析与方程}
\cfoot{}

\begin{document}
%\maketitle

%\fancypage{%
%\setlength{\fboxsep}{13pt}%
%\setlength{\fboxrule}{0.8pt}%

\begin{center}
{\erhao \CJKfamily{kd}{第一届Xionger网络数学竞赛} }\\
\vspace{0.3cm}
{\sanhao \textbf{分析与方程部分试题解答} }\\
\vspace{0.2cm}
2018年6月8日}
%{\sanhao \textbf{解答：Eufisky (Xiongge)}}\\
\end{center}
%\textbf{分析与代数：}
%满分为100分，考试时间为120分钟，答案必须写在答题纸上\\
%\vspace*{-0.05cm}
\rule{\textwidth}{0.5mm}
%\rule{\textwidth}{0.4pt}
%\vspace*{-0.05cm}
%\boxed{\text{\textbf{注意：答案必须写在答题纸上，写在试卷或草稿纸上均无效。}}}

\begin{enumerate}
%\renewcommand{\labelenumi}{\textbf{{\theenumi}.}}
% 重定义第一级计数显示
\renewcommand{\theenumi}{\textbf{\arabic{enumi}.}}
\renewcommand{\labelenumi}{\theenumi}

% 重定义第二级计数显示
\renewcommand{\theenumii}{\alph{enumii}}
\renewcommand{\labelenumii}{(\theenumii)}

\item Suppose $S_n=a_1+\cdots+a_n$, and $T_n=S_1+\cdots+S_n$. If $\lim\limits_{n\to\infty} \frac{1}{n}T_n=s$ and $na_n$ is bounded, then $\lim\limits_{n\to\infty} S_n=s.$
\begin{proof}
We may assume that $\lim\limits_{n\to\infty} \frac{1}{n}T_n=0$, or we can replace $a_1$ by $a_1-s$. For $\forall \varepsilon>0$, there is a large number $N$, $\forall n\ge N$, we have $|T_n|\le \varepsilon n.$ Since
$$T_{n+k}-T_n=S_{n+1}+\cdots+S_{n+k}$$
$$=kS_n+(ka_{n+1}+(k-1)a_{n+2}+\cdots+2a_{n+k-1}+a_{n+k}),$$
we have
$$kS_n=T_{n+k}-T_n-(ka_{n+1}+(k-1)a_{n+2}+\cdots+2a_{n+k-1}+a_{n+k}).$$
Suppose $|na_n|\le C$, then
$$k|S_n|\le (2n+k)\varepsilon+C\left(\frac{k}{n+1}+\frac{k-1}{n+2}+\cdots+\frac{1}{n+k}\right)\le (2n+k)\varepsilon+C\frac{k^2}{n},$$
or equivalent,
$$|S_n|\le \left(\frac{2n}{k}+1\right)\varepsilon+C\frac{k}{n}.$$
Take $k=[\sqrt{\varepsilon}n]>\frac{1}{2}\sqrt{\varepsilon}n$, then we have
$$|S_n|\le \left(\frac{4}{\sqrt{\varepsilon}}+1\right)\varepsilon+C\sqrt{\varepsilon}=\left(4+\sqrt{\varepsilon}+C\right)\sqrt{\varepsilon}.$$
\end{proof}

%\begin{solution}

%\end{solution}

\item Suppose $f$ is a real value function on $\mathbb{R}$, and $f(x+y)=f(x)+f(y)$ for $\forall x, y\in \mathbb{R}$. If the set $\{(x,f(x)): x\in\mathbb{R}\}$ is not dense in $\mathbb{R}^2$, then $f$ is continuous.
\begin{proof}
In fact, we have $f(x)\equiv f(1)x$. It is not hard to see $f(rx)=rf(x)$ for $x\in\mathbb{R}, r\in \mathbb{Q}$. We denote $c=f(1)$, then $f(r)=cr$ for $r\in \mathbb{Q}$. If $f(x)\equiv cx$ is false, then there exists a number $x_0\in \mathbb{R}-\mathbb{Q}$, s.t. $y_0=f(x_0)\ne cx_0$. Then we have
$$f(x_0s+r)=y_0s+cr=c(x_0s+r)+(y_0-cx_0)s, \forall s, t\in \mathbb{Q}.$$
Fix $y\in\mathbb{R}$. For $\forall \varepsilon>0$, there is a number $s\in\mathbb{Q}$, s.t. $|s(y_0-cx_0)-y|<\varepsilon$, then $|s|< \frac{|y|+\varepsilon}{|y_0-cx_0|}\le \frac{|y|+1}{|y_0-cx_0|}=C$. Take $r\in\mathbb{Q}$ s.t. $|x_0+r|<\varepsilon$, we have
$$f(x_0+r)=c(x_0+r)+(y_0-cx_0),$$
and
$$f(s(x_0+r))=sc(x_0+r)+s(y_0-cx_0),$$
then $|s(x_0+r)|\le C\varepsilon$ and
$$|f(s(x_0+r))-y|\le |sc(x_0+r)|+|s(y_0-cx_0)-y|< (Cc+1)\varepsilon.$$
Fix $(x,y)\in\mathbb{R}^2$. For $\forall \varepsilon>0$, there exists a number $a\in \mathbb{R}$ s.t. $|a|<\varepsilon$, and $|f(a)-(y-cx)|< \varepsilon$. Take $r\in \mathbb{Q}$ s.t. $|x-r|<\varepsilon$, we have
$$f(a+r)=f(a)+cr=f(a)+cx+c(r-x)$$
then $|(a+r)-x|<2\varepsilon$ and
$$|f(a+r)-y|=|f(a)-(y-cx)|+|c(r-x)|< (1+c)\varepsilon.$$
\end{proof}

\item Suppose $\Omega\subset \mathbb{C}$ is a domain, and $u_n=Re f_n$, where $f_n\in H(\Omega)$. If $\{u_n\}$ is uniform convergence on arbitrary compact subset of $\Omega$, and $\{f_n(z_0)\}$ is convergence for some $z_0\in \Omega$. Prove that $\{f_n\}$ is uniform convergence on arbitrary compact subset of $\Omega$.
\begin{proof}
Since the compact set has a finite open covering property, we only need to consider the case $\Omega=\mathbb{D}$. We can obtain the conclusion by the Borel-Carath$\acute{\text{e}}$odory lemma: Suppose $f\in H(\mathbb{D})$. Let $M(r)=\max\limits_{|z|=r} |f(z)|$, $A(r)=\max\limits_{|z|=r} Re f(z)$, then for $0<r<R<1$, we have
$$M(r)\le \frac{2r}{R-r}A(R)+\frac{R+r}{R-r}|f(0)|.$$
\end{proof}

\item  Suppose $\Omega\subset \mathbb{C}$ is a convex domain, and $f\in H(\Omega)$. If $Re f'(z)\ge 0$ for $\forall z\in\Omega$ and $f$ is not constant function, then $f$ is injective.
\begin{proof}
Consider $e^{-f'(z)}\in H(\Omega)$, then $|e^{-f'(z)}|=e^{-Re f'(z)}\le 1$. If $Re f'(z_0)=0$ for some $z_0\in\Omega$, then $e^{-f'(z)}=const$, or $f'(z)=const$ by the maximum principle. That is to say $f(z)=cz$ with $c\ne 0$, then $f$ is injective. If $Re f'(z)> 0$ for $\forall z\in\Omega$, then for $\forall z_1, z_2\in\Omega$ with $z_1\ne z_2$, we have
$$f(z_2)-f(z_1)=\int_{z_1}^{z_2}f'(z)dz=(z_2-z_1)\int_{0}^{1} f'(z_1+t(z_2-z_1))dt,$$
then
$$Re\frac{f(z_2)-f(z_1)}{z_2-z_1}=\int_{0}^{1} Re f'(z_1+t(z_2-z_1))dt>0.$$
\end{proof}

%\begin{enumerate}[itemsep=0pt,parsep=0pt,label=(\arabic*)]
%\item
%\end{enumerate}

%\begin{newproof}

%\end{newproof}

\item  Prove the linear span of $t^ne^{-t}, n=0,1,2,\cdots$ is dense in $L^2(0,\infty)$.
\begin{proof}
Let $M$ be the closed linear span of $t^ne^{-t}, n=0,1,2,\cdots$. Take any $\varphi\in M^{\bot}$, we have
$$\int_{0}^{\infty} t^ne^{-t}\varphi(t)dt=0, n=0,1,2,\cdots.$$
Let $z$ be a complex with $\Im\ z>-1$, and
$$f(z)=\int_{0}^{\infty} e^{izt}e^{-t}\varphi(t)dt.$$
Since $|\frac{e^z-1}{z}|\le C\max\{1,e^{|z|}\}$, we have
$$f'(z)=\lim\limits_{h\to 0}\frac{f(z+h)-f(z)}{h}=\lim\limits_{h\to 0} \int_{0}^{\infty} \frac{e^{iht}-1}{h}e^{izt}e^{-t}\varphi(t)dt$$
$$=\int_{0}^{\infty} ite^{izt}e^{-t}\varphi(t)dt$$
by the dominated convergence theorem. That means $f$ is analytic. Similarly, we have
$$f^{(n)}(z)=\int_{0}^{\infty} i^nt^ne^{izt}e^{-t}\varphi(t)dt.$$
Since $f^{(n)}(0)=i^n\int_{0}^{\infty} t^ne^{-t}\varphi(t)dt=0, n=0,1,2,\cdots$, we have $f(z)\equiv 0$, that means $e^{izt}e^{-t}\in M$. According to the Weierstrass approximation theorem, every continuous periodic function $h(t)$ is the uniform limit of trigonometric polynomials, we can get $h(t)e^{-t}\in M$. Let $g(t)$ be a continuous function with compact support, and $g_1(t)=g(t)e^t$. Denote by $h(t)$ a $T$ periodic function such that
$$h(t)\equiv g_1(t), t\in [0,T],$$
where $T$ is large enough so that the support of $g_1(t)$ is contained in the interval $[0,T]$. Then
$$|g_1(t)-h(t)|\le ||g_1||_{L^{\infty}}\chi_{(T,\infty)}(t),$$
so that
$$|g(t)-h(t)e^{-t}|\le ||g_1||_{L^{\infty}}e^{-t}\chi_{(T,\infty)}(t).$$
Let $T\to\infty$, we can get $g(t)\in M$. Since the set of all continuous functions with compact support is dense in $L^2(0,\infty)$, we have $M=L^2(0,\infty)$.
\end{proof}

\item Let  $A$ is a unital commutative Banach algebra that is generated by $\{1,x\}$ for some $x\in A$. Then the complement set of $\sigma(x)$ is connected.
\begin{proof}
Let us decompose $\sigma(x)^c$ into its connected components, obtaining an unbounded component $\Omega_{\infty}$ together with a sequence of holes $\Omega_1, \Omega_2, \cdots,$
$$\sigma(x)^c=\Omega_{\infty}\cup\Omega_1\cup\Omega_2\cup\cdots.$$
Let $\Omega=\Omega_1\cup\Omega_2\cup\cdots$. If $\sigma(x)^c$ is not connected, the $\Omega\ne \emptyset$. Suppose $\lambda\in\Omega$, then for arbitrary polynomial $p(z)$, since $p(z)$ is analytic, we have
$$|p(\lambda)|\le \max_{z\in \sigma(x)} |p(z)|=\max\limits_{\omega\in Sp(A)} |\omega(p(x))|\le ||p(x)||$$
by the maximum principle and Gelfand theorem. If we defind
$$\omega: p(x)\mapsto p(\lambda),$$
then $\omega$ is bounded on $\{p(x)\}$. Since $\{p(x)\}$ is dense in $A$, $\omega$ have unique extension on $A$, and $\omega(xy)=\omega(x)\omega(y)$, that means $\omega\in Sp(A)$. Then $\lambda=\omega(x)\in\sigma(x)$, it is a contradiction.
\end{proof}

\item Suppose $X$ is a compact  Hausdorff space. $\Omega$ is a family of colsed connected subset of $X$, and $\Omega$ is totally order with respect to inclusion relation. Then $Y=\cap\{A: A\in\Omega\}$ is connected.
\begin{proof}
If $Y$ is not connected, then are open set $B$ and $C$, with $B\cap C=\emptyset$, $B\cap Y\ne \emptyset$ and $C\cap Y\ne \emptyset$. Consider the set $Y_1=\cap\{A-(B\cup C): A\in\Omega\}$, then $Y_1=Y-(B\cup C)=\emptyset$. Since $A$ is connected, if $A-(B\cup C)=\emptyset$, or $A\subset B\cup C$, then $A\subset B$, or $A\subset C$, it is impossible. Thus $A-(B\cup C)\ne\emptyset$. Since $A-(B\cup C)$ is compact, and finite intersection is not empty, then $Y_1\ne\emptyset$. It is a contradiction.
\end{proof}

\item Suppose the measurable set $A\subset \mathbb{R}$ with $0<m(A)<\infty$. Let $f(x,r)=m(A\cap[x-r,x+r])/2r$, then there exists $x\in\mathbb{R}$ s.t.
$$0<\liminf\limits_{r\to 0_+}f(x,r)\le \limsup\limits_{r\to 0_+}f(x,r)<1.$$
\begin{proof}
Since $0<m(A)<\infty$, there are interval $I_1, I_2$ with $|I_1|=|l_2|=2r_0$ s.t. $m(A\cap I_1)>\frac{1}{2}|I_1|$, $m(A\cap I_2)<\frac{1}{2}|I_2|$.
Since $f(x,r)$ is continuous about $x$, there exists $x_0$ with $f(x_0,r_0)=\frac{1}{2}$. Since we have
$$m(A\cap [x_0-r_0,x_0+r_0])=m(A\cap [x_0-r_0, x_0])+m(A\cap [x_0, x_0+r_0])=r_0,$$
there exists $x_1\in [x_0-\frac{r_0}{2},x_0+\frac{r_0}{2}]$ with $f(x_1,\frac{r_0}{2})=\frac{1}{2}$. Or equivalently, there exists $x_1\in\mathbb{R}$ with
$$|x_1-x_0|\le\frac{r_0}{2}, f(x_1,\frac{r_0}{2})=\frac{1}{2}.$$
Similarly, there exists $x_n\in\mathbb{R}$ with
$$|x_n-x_{n-1}|\le\frac{r_0}{2^n},\quad f\left(x_n,\frac{r_0}{2^n}\right)=\frac{1}{2}.$$
Let $x=\lim\limits_{n\to\infty} x_n$, then $|x-x_n|=\left|\sum\limits_{k=n+1}^{\infty} (x_k-x_{k-1})\right|\le \sum\limits_{k=n+1}^{\infty} |x_k-x_{k-1}|\le\frac{r_0}{2^n}$. For any $r<r_0$, there exists unique $N\ge 1$ s.t. $\frac{r_0}{2^N}\le r<\frac{r_0}{2^{N-1}}$. Then we have
$$\left[x_{N+1}-\frac{r_0}{2^{N+1}}, x_{N+1}+\frac{r_0}{2^{N+1}}\right]\subset\left[x-\frac{r_0}{2^N},x+\frac{r_0}{2^N}\right]\subset[x-r,x+r],$$
and
\begin{align*}
f(x,r)&=\frac{m\left( A\cap \left[ x-r,x+r \right] \right)}{2r}\ge \frac{m\left( A\cap \left[ x_{N+1}-\frac{r_0}{2^{N+1}},x_{N+1}+\frac{r_0}{2^{N+1}} \right] \right)}{2\times\frac{r_0}{2^{N-1}}}\\
&=\frac{1}{4}f\left( x_{N+1},\frac{r_0}{2^{N+1}} \right) =\frac{1}{8}.
\end{align*}

On the other hand, we have
\begin{align*}
m\left( A\cap \left[ x-r,x+r \right] \right) &\le m\left( A\cap \left[ x_{N+1}-\frac{r_0}{2^{N+1}},x_{N+1}+\frac{r_0}{2^{N+1}} \right] \right)
\\
&+m\left( \left[ x-r,x+r \right] -\left[ x_{N+1}-\frac{r_0}{2^{N+1}},x_{N+1}+\frac{r_0}{2^{N+1}} \right] \right)
\\
&=\frac{r_0}{2^{N+1}}+2r-\frac{2r_0}{2^{N+1}}=2r-\frac{r_0}{2^{N+1}}\le 2r-\frac{1}{4}r=\frac{7}{4}r,
\end{align*}
then
$$f(x,r)=\frac{m(A\cap[x-r,x+r])}{2r}\le\frac{7}{8}.$$
That is to say
$$\frac{1}{8}\le\liminf\limits_{r\to 0_+}f(x,r)\le \limsup\limits_{r\to 0_+}f(x,r)\le\frac{7}{8}.$$
\end{proof}

\item Suppose $\{f_n\}_{n=1}^{\infty}$ is a bounded sequence in $L^p$ with $1\le p<\infty$. If $f_n\to f$ a.e., then $f\in L^p$ and
$$\lim\limits_{n\to\infty}\int |f_n|^p-|f_n-f|^p=\int |f|^p.$$
\begin{proof}
We denote $M=\mathop{\sup}_{n\ge 1}\int |f_n|^p<\infty$.Since $f_n\to f$ a.e., we have $|f_n|^p\to|f|^p$ a.e. and by Fatou Lemma
$$\int |f|^p\le\mathop{\underline{\lim}}\limits_{n\to\infty}\int |f_n|^p\le M<\infty,$$
that is to say $f\in L^p$. For $\forall a,b\ge 0$, we have
$$|a^p-b^p|=p\xi^{p-1}|a-b|\le p\max\{a,b\}^{p-1}|a-b|$$
by Lagrange Mean Value Theorem, where $\xi$ is a real number between $a$ and $b$. Then we obtain
$$||f_n|^p-|f_n-f|^p|\le p\max\{|f_n|,|f_n-f|\}^{p-1}|f|.$$
Fixed $\varepsilon>0$. Suppose $A$ is a measurable set with $m(A)<\infty$, and the follow inequality holds
$$\int_{A^c} |f|^p\le \varepsilon.$$
There is a $\delta>0$ such that
$$\int_B |f|^p<\varepsilon \text{ whenever } m(B)<\delta$$
by absolute continuity. Since $f_n\to f$ a.e. on $A$ and $m(A)<\infty$, we can find a measurable subset $a\subset A$ such $m(A\setminus a)<\delta$ and $f_n\to f$ uniformly on $a$ by Egorov Theorem. Then we have
$$\int_a |f_n|^p-|f_n-f|^p\to \int_a |f|^p,$$
as $n\to\infty$, and
$$\int_{a^c} |f|^p=\int_{A\setminus a} |f|^p+\int_{A^c} |f|^p< 2\varepsilon.$$
Since the function $\max\{|f_n|,|f_n-f|\}^{p-1}\in L^{p'}$, and $$||\max\{|f_n|,|f_n-f|\}^{p-1}||_{p'}=||\max\{|f_n|,|f_n-f|\}||_p^{p-1}\le (3M^p)^{\frac{1}{p'}}.$$
We have
\begin{align*}
\int_{a^c} ||f_n|^p-|f_n-f|^p|&\le p\int_{a^c}\max\{|f_n|,|f_n-f|\}^{p-1}|f|\\
&\le p(3M)^{\frac{1}{p'}}(\int_{a^c} |f|^p)^{\frac{1}{p}}
\le p(3M)^{\frac{1}{p'}}(2\varepsilon)^{\frac{1}{p}},
\end{align*}
by H\"{o}lder inequality. Thus we obtain
$$\mathop{\overline{\lim}}\limits_{n\to\infty}\left|\int |f_n|^p-|f_n-f|^p-\int |f|^p\right|\le (1+p(3M)^{\frac{1}{p'}})(2\varepsilon)^{\frac{1}{p}}.$$
\end{proof}

\item Suppose $D_n(t)$ are the Dirichlet kernels, and $F_N(t)$ is the $N$-th Fej$\acute{\text{e}}$r kernel given by
$$F_N(t)=\frac{D_0(t)+\cdots+D_{N-1}(t)}{N}.$$
Let $L_N(t)=\min\left(N,\frac{\pi^2}{Nt^2}\right)$. Prove
$$F_N(t)=\frac{1}{N}\frac{1-\cos Nt}{1-\cos t}\le L_N(t)$$
and $\int_{\mathbb{T}} L_N(t)dt\le 4\pi$. If $f\in L^1(\mathbb{T})$ and the $N$-th Ces$\grave{\text{a}}$ro mean of Fourier series is
$$\sigma_N(f)(x)=\frac{S_0(f)(x)+\cdots+S_{N-1}(f)(x)}{N},$$
then $\sigma_N(f)(x)\to f(x)$ for every $x$ in the Lebesgue set of $f$.
\begin{proof}
Since $D_N(t)=\sum\limits_{n=-N}^{N}e^{int}=\frac{\sin(N+\frac{1}{2})t}{\sin\frac{t}{2}}$, we have
$$F_N(t)=\frac{1}{N}\frac{\sin^2\frac{Nt}{2}}{\sin^2\frac{t}{2}}=\frac{1}{N}\frac{1-\cos Nt}{1-\cos t}.$$
Since $|D_N(t)|\le 2N+1$, we can get $F_N(t)\le N$. For $0<x<\frac{\pi}{2}$, we have $\sin x\ge\frac{2}{\pi}x$, then
$$F_N(t)=\frac{1}{N}\frac{\sin^2\frac{Nt}{2}}{\sin^2\frac{t}{2}}\le\frac{1}{N}\frac{1}{\sin^2\frac{t}{2}}\le\frac{\pi^2}{Nt^2}.$$
That mens $F_N(t)\le L_N(t)$. And
$$\int_{\mathbb{T}} L_N(t)dt=2\int_0^{\pi}L_N(t)dt=2\int_0^{\frac{\pi}{N}}Ndt+2\int_{\frac{\pi}{N}}^{\pi}\frac{\pi^2}{Nt^2}dt=4\pi-\frac{2\pi}{N} \le 4\pi.$$
Since $\int_{\mathbb{T}}F_N(t)=1$ and $F_N(t)\le L_N(t)$, we can get $\{F_N(t)\}$ is an approximation to the identity, then $\sigma_N(f)(x)=(f*F_N)(x)\to f(x)$ for every $x$ in the Lebesgue set of $f$.
\end{proof}

\item Suppose the sequence $\{a_n\}$ satisfying $a_{n+1}=(4n-2)a_n+a_{n-1}$. Prove that $\{a_n\}$ is convergence if and only if
$$(e-1)a_0+(e+1)a_1=0.$$
\begin{proof}
If $\{a_n\}$ is convengence and not vanishing, then it is obvious that $a_na_{n+1}<0$. We can assume $a_0>0$, $a_1<0$, then $a_{2n}>0$, $a_{2n+1}<0$. Let $b_n=4n-2$ and
$$a_n=p_{n-2}a_1+q_{n-2}a_0, n\ge 2.$$
Since $a_2=b_1a_1+a_0$, and $a_3=(1+b_1b_2)a_1+b_2a_0$, we can get
$$p_0=b_1, p_1=1+b_1b_2, q_0=1, q_1=b_2.$$
Since $a_{n+2}=b_{n+1}a_{n+1}+a_n=b_{n+1}(p_{n-1}a_1+q_{n-1}a_0)+(p_{n-2}a_1+q_{n-2}a_0)=(b_{n+1}p_{n-1}+p_{n-2})a_1+(b_{n+1}q_{n-1}+q_{n-2})a_0$, we can get
$$p_n=b_{n+1}p_{n-1}+p_{n-2}, q_n=b_{n+1}q_{n-1}+q_{n-2}.$$
That is to say
$$\frac{p_n}{q_n}=\left[b_1,b_2,\cdots,b_{n+1}\right]=b_1+\cfrac{1}{b_2+\cfrac{1}{\cdots+\cfrac{1}{b_{n+1}}}},$$
and we have $\frac{p_n}{q_n}\to [b_1,b_2,\cdots]$ as $n\to\infty$. Since
$$a_{2n+1}=p_{2n-1}a_1+q_{2n-1}a_0<0, a_{2n+2}=p_{2n}a_1+q_{2n}a_0>0,$$
we have
$$\frac{p_{2n}}{q_{2n}}(-a_1)<a_0<\frac{p_{2n-1}}{q_{2n-1}}(-a_1),$$
Let $n\to\infty$, we can get
$$a_0+[b_1,b_2,\cdots]a_1=0.$$
On the contrary, if $a_0+[b_1,b_2,\cdots]a_1=0$, since $\left|[b_1,b_2,\cdots]-\frac{p_n}{q_n}\right|<\frac{1}{q_nq_{n+1}}$, we can get
$$|a_n|=|a_1|\cdot\left|p_{n-2}-q_{n-2}[b_1,b_2,\cdots]\right|\le\frac{|a_1|}{q_{n-1}}\to 0.$$
Since
$$\frac{e-1}{e+1}=[0,2,6,10,\cdots]=\cfrac{1}{2+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{\cdots}}}},$$
we can get $[b_1,b_2,\cdots]=\frac{e+1}{e-1}$, then $a_0+[b_1,b_2,\cdots]a_1=0$ is equivalent to
$$(e-1)a_0+(e+1)a_1=0.$$
\end{proof}

\item Suppose $\{x_n\}$ satisfying $x_1=1$, $x_{n+1}=x_n+\frac{1}{S_n}$, where $S_n=x_1+\cdots+x_n$. Prove that\\
(a) $x_n^2-2\ln S_n$ is increasing and $x_n^2-2\ln S_{n-1}$ is decreasing for $n\ge 2$.\\
(b) $x_n^2-2\ln n-\ln\ln n$ is convengence.\\
(c) $\lim\limits_{n\to\infty} \cfrac{\ln n}{\ln \ln n}\left(\cfrac{x_n}{\sqrt{2\ln n}}-1\right)=\frac{1}{4}$.
\begin{proof}
For $n\ge 2$, we have
\begin{align*}
&\left( x_{n+1}^{2}-\text{2}\ln S_n \right) -\left( x_{n}^{2}-\text{2}\ln S_{n-1} \right) =\left( x_n+\frac{1}{S_n} \right) ^2-x_{n}^{2}+\text{2}\ln \frac{S_{n-1}}{S_n}
\\
&=\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}+\text{2}\ln \left( 1-\frac{x_n}{S_n} \right) <\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}-2\left( \frac{x_n}{S_n}+\frac{x_{n}^{2}}{2S_{n}^{2}} \right) =\frac{1-x_{n}^{2}}{2S_{n}^{2}}\le 0.
\end{align*}

Similarly,
\begin{align*}
&\left( x_{n+1}^{2}-\text{2}\ln S_{n+1} \right) -\left( x_{n}^{2}-\text{2}\ln S_n \right) =\left( x_n+\frac{1}{S_n} \right) ^2-x_{n}^{2}-\text{2}\ln \frac{S_{n+1}}{S_n}
\\
&=\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}-\text{2}\ln \left( 1+\frac{x_{n+1}}{S_n} \right) >\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}-2\left( \frac{x_{n+1}}{S_n}-\frac{x_{n+1}^{2}}{2S_{n}^{2}}+\frac{x_{n+1}^{3}}{3S_{n}^{3}} \right)
\\
&=\frac{x_{n+1}^{2}-1}{S_{n}^{2}}-\frac{2}{3}\frac{x_{n+1}^{3}}{S_{n}^{3}}=\frac{x_{n+1}^{2}}{S_{n}^{2}}\left( 1-\frac{1}{x_{n+1}^{2}}-\frac{2x_{n+1}}{3S_n} \right).
\end{align*}
Since $x_1=1, x_2=2$, and $S_1=1, S_2=3$, we have
$$x_{n+1}-S_n=x_n+\frac{1}{S_n}-S_n=\frac{1}{S_n}-S_{n-1}\le \frac{1}{S_2}-S_1=-\frac{1}{2}<0,$$
that means $\frac{2x_{n+1}}{3S_n}\le\frac{2}{3}$, then
$$1-\frac{1}{x_{n+1}^2}-\frac{2x_{n+1}}{3S_n}>\frac{1}{3}-\frac{1}{x_{n+1}^2}>0.$$
That implys (a).\\
Since $x_n^2-2\ln S_n$ is increasing, we can get $x_n^2-2\ln S_n\ge x_2^2-2\ln S_2=4-2\ln 3>1$. Since $x_n\ge 1$, we have $S_n\ge n$, then
$$x_n^2\ge 1+2\ln S_n\ge 1+2\ln n.$$
Since $x_n^2-2\ln S_{n-1}$ is decreasing, we can get $x_n^2-2\ln S_n\le x_{n+1}^2-2\ln S_n\le x_2^2-2\ln S_2=4$, then
$$x_n^2\le 4+2\ln S_n.$$
Since $S_n\ge n$, we can get $x_n\le 1+1+\frac{1}{2}+\cdots+\frac{1}{n-1}\le 2+\ln n$, then $S_n\le nx_n\le n(2+\ln n)$, and
$$x_n^2\le 4+2\ln n+2\ln(2+\ln n).$$
Thus, it is obvious that
$$\frac{x_n}{\sqrt{\ln n}}\to\sqrt{2},$$
and we have
$$\frac{S_n}{n\sqrt{\ln n}}\to \sqrt{2}$$
by Stolz formula. Since $x_n^2-2\ln S_n\le 4$, we can get $x_n^2-2\ln S_n$ is convengence, then
$$x_n^2-2\ln n-\ln\ln n=x_n^2-2\ln S_n+2\ln\frac{S_n}{n\sqrt{\ln n}}$$
is convengence. That implys (b).\\
Since $x_n^2=2\ln n+\ln\ln n+a+o(1)$, we can get
$$\frac{x_n^2}{2\ln n}-1=\frac{\ln \ln n}{2\ln n}+\frac{a+o(1)}{2\ln n},$$
then
\begin{align*}
\cfrac{\ln n}{\ln \ln n}\left(\cfrac{x_n}{\sqrt{2\ln n}}-1\right)&=\left(\cfrac{x_n}{\sqrt{2\ln n}}+1\right)^{-1}\cfrac{\ln n}{\ln \ln n}\left(\frac{x_n^2}{2\ln n}-1\right)\\
&\to \frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}.
\end{align*}
Moreover, we have
$$\lim\limits_{n\to\infty} \ln \ln n\left(\cfrac{\ln n}{\ln \ln n}\left(\cfrac{x_n}{\sqrt{2\ln n}}-1\right)-\frac{1}{4}\right)=\frac{a}{4}.$$
\end{proof}

\item Suppose $f\in C[0,\infty)$ and for $\forall a\ge0$, we have
$$\lim\limits_{x\to\infty} f(x+a)-f(x)=0.$$
Then there exist $g\in C[0,\infty)$ and $h\in C^1[0,\infty)$ with $f=g+h$, such that
$$\lim\limits_{x\to\infty} g(x)=0, \text{ } \lim\limits_{x\to\infty} h'(x)=0.$$
\begin{proof}
In fact, $f$ is uniformly continuous. Otherwise, there are two sequences $\{x_n\}_{n=1}^{\infty}$, $\{y_n\}_{n=1}^{\infty}$ and a positive $\varepsilon$, such that
$$x_n, y_n\to\infty, |x_n-y_n|\to 0, |f(x_n)-f(y_n)|\ge\varepsilon_0$$
as $n\to\infty$. Consider the functions
$$\varphi_n(x)=f(x_n+x)-f(x_n)$$
and
$$\phi_n(x)=f(y_n+x)-f(y_n)$$
defined on the interval [0,1]. Then we have
$$\varphi(x), \phi(x)\to 0$$
as $n\to\infty$ due to $\lim\limits_{x\to\infty} f(x+a)-f(x)=0$. For $\forall 0<\varepsilon<\cfrac{1}{2}$, there is a set $A_{\varepsilon}\in[0,1]$ such that $m([0,1]\setminus A_{\varepsilon})<\varepsilon$ and $\varphi_n, \phi_n\to 0$ uniformly on $A_{\varepsilon}$ by Egorov Theorem. Take a integer $N$ such that $\forall n\ge N$ and $\forall x\in A_{\varepsilon}$, we have $|x_n-y_n|<1-2\varepsilon$ and
$$|\varphi_n(x)|\le\cfrac{\varepsilon_0}{3},\qquad |\phi_n(x)|\le\cfrac{\varepsilon_0}{3}.$$
Since $m((x_n+A_{\varepsilon})\cap(y_n+A_{\varepsilon}))=m(x_n+A_{\varepsilon})+m(y_n+A_{\varepsilon})-m((x_n+A_{\varepsilon})\cup(y_n+A_{\varepsilon}))\ge 2(1-A_{\varepsilon})-(1+|x_n-y_n|)=1-2\varepsilon-|x_n-y_n|>0$, there is a point $x\in(x_n+A_{\varepsilon})\cap(y_n+A_{\varepsilon})$. We have $x-x_n,x-y_n\in A_{\varepsilon}$, and then
$$|\varphi(x-x_n)|=|f(x)-f(x_n)|\le\cfrac{\varepsilon_0}{3}, \quad |\phi(x-y_n)|=|f(x)-f(y_n)|\le\cfrac{\varepsilon_0}{3},$$
thus $|f(x_n)-f(y_n)|\le \cfrac{2}{3}\varepsilon_0$, it is a contradiction.

Let $h(x)=\int_{x}^{x+1} f(t)dt$, and $g(x)=f(x)-h(x)$, then we have
$$h'(x)=f(x+1)-f(x)\to 0$$
as $x\to\infty$. Since $f$ is uniformly continous, there is positive $M$ such $\forall x,y\ge 0$ with $|x-y|\le 1$, we have $|f(x)-f(y)|\le M$. Since $f(x)-f(x+t)\to 0$ as $x\to\infty$ and $|f(x)-f(x+t)|\le M$ for $\forall t\in [0,1]$, by DCT (Dominated convergence theorem), we have
$$g(x)=\int_{0}^{1} f(x)-f(x+t)\to 0 \text{ as } n\to\infty.$$
\end{proof}

\end{enumerate}

\end{document}

## 向老师的题目

$\int_0^1{\frac{1}{1+a^2x^2}\left[\left(1-\frac{x}{2}\right)\ln\frac{1+x}{1-x}+\frac{\pi^2x^2}{4}\right]^{-1}\textrm{d}x}.$

\begin{align*}&\hspace{0.5cm}\int_0^1{\frac{1}{1+a^2x^2}\left[\left(1-\frac{x}{2}\right)\ln\frac{1+x}{1-x}+\frac{\pi^2x^2}{4}\right]^{-1}\textrm{d}x}\\&=\int_0^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\left(\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2x^2}{4}}\right)\textrm{d}x}\\&=\frac{1}{2}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\left(\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2x^2}{4}}\right)\textrm{d}x}\end{align*}

\begin{align*}&\int_{ - \infty  - \frac{\pi }{2}{\rm{i}}}^{\infty  - \frac{\pi }{2}{\rm{i}}} {f\left( z \right){\rm{d}}z}  - \int_{ - \infty  + \frac{\pi }{2}{\rm{i}}}^{\infty  + \frac{\pi }{2}{\rm{i}}} {f\left( z \right){\rm{d}}z} \\&= 2\pi {\rm{i}}\left( {{\rm{Res}}\left[ {f\left( z \right),z = 0} \right] + \left( {{\rm{Res}}\left[ {f\left( z \right),{\rm{i}} \cdot \arctan \left( a \right)} \right] + {\rm{Res}}\left[ {f\left( z \right),z =  - {\rm{i}} \cdot \arctan \left( a \right)} \right]} \right)} \right)\\&+ \pi {\rm{i}}\left( {{\rm{Res}}\left[ {f\left( z \right),z = \frac{\pi }{2}{\rm{i}}} \right] + {\rm{Res}}\left[ {f\left( z \right),z =  - \frac{\pi }{2}{\rm{i}}} \right]} \right)\\&= 2\pi {\rm{i}}\left( {\frac{3}{{{a^2}}} - \frac{a}{{2\left( {a - \arctan \left( a \right)} \right)}} - \frac{a}{{2\left( {a - \arctan \left( a \right)} \right)}}} \right) + \pi {\rm{i}}\left( {1 + 1} \right)\\&= 2\pi {\rm{i}}\left( {\frac{3}{{{a^2}}} - \frac{{\arctan \left( a \right)}}{{a - \arctan \left( a \right)}}} \right).\end{align*}

\begin{align*}&\int_{-\infty -\frac{\pi}{2}\textrm{i}}^{\infty -\frac{\pi}{2}\textrm{i}}{f\left(z\right)\textrm{d}z}-\int_{-\infty +\frac{\pi}{2}\textrm{i}}^{\infty +\frac{\pi}{2}\textrm{i}}{f\left(z\right)\textrm{d}z}=\int_{-\infty}^{\infty}{f\left(x-\frac{\pi}{2}\textrm{i}\right)\textrm{d}x}-\int_{-\infty}^{\infty}{f\left(x+\frac{\pi}{2}\textrm{i}\right)\textrm{d}x}\\&=\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\coth x-\left(x-\frac{\pi}{2}\textrm{i}\right)^2}\textrm{d}x}-\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\coth x-\left(x+\frac{\pi}{2}\textrm{i}\right)^2}\textrm{d}x}\\&=-\pi\textrm{i}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}=2\pi\textrm{i}\left(\frac{3}{a^2}-\frac{\arctan\left(a\right)}{a-\arctan\left(a\right)}\right).\end{align*}

\begin{align*}\int_0^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}&=\frac{1}{2}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}\\&=\frac{\arctan\left(a\right)}{a-\arctan\left(a\right)}-\frac{3}{a^2}\end{align*}

$\int_0^1{x^{20}\left[ \left( 1-\frac{x}{2}\ln \frac{1+x}{1-x} \right) ^2+\frac{\pi ^2x^2}{4} \right] ^{-1}\text{d}x}=\frac{5588512716806912356}{374010621408251953125}$

$\sum_{n=1}^\infty\frac{\mathrm{Si}(n\pi)}{n^3}$

\begin{align*}\text{Si}\left( n\pi \right) &=\int_0^{n\pi}{\frac{\sin t}{t}\text{d}t}=\int_0^{\pi}{\frac{\sin nx}{x}\text{d}x}=\int_0^{\pi}{\sin nx\text{d}\left( \ln x \right)}=-n\int_0^{\pi}{\cos nx\ln x\text{d}x}\\&=-n\int_0^{\pi}{\cos nx\text{d}\left( x\ln x-x \right)}=n\left[ \left( -1 \right) ^{n-1}\left( \pi \ln \pi -\pi \right) -n\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x} \right]\end{align*}

$\sum_{n=1}^{\infty}{\frac{\text{Si}(n\pi)}{n^3}}=\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n^2}\left( \pi \ln \pi -\pi \right)}-\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}$

\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\text{d}\left( \frac{1}{2}x^2\ln x-\frac{3}{4}x^2 \right)}}\\&=\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \cos nx\text{d}x}}.\end{align*}

$\widetilde{f}\left( x \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n\cos nx}$

$\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n}=f(0)=0$

$\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \text{d}x}}=\frac{\pi}{2}\sum_{n=1}^{\infty}{a_n}=-\frac{\pi}{4}a_0=\frac{\pi ^3}{12}\ln \pi -\frac{11}{72}\pi ^3$

$\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^3}}=\frac{\pi ^2}{12}\left( \pi \ln \pi -\pi \right) -\left( \frac{\pi ^3}{12}\ln \pi -\frac{11}{72}\pi ^3 \right) =\frac{5\pi ^3}{72}$

$\sum_{n=1}^{\infty}{\left( -1 \right) ^n\frac{\text{Si}\left( n\pi \right)}{n^3}}=-\frac{\pi ^2}{6}\left( \pi \ln \pi -\pi \right) -\left( \frac{2\pi ^3}{9}-\frac{\pi ^3}{6}\ln \pi \right) =-\frac{\pi ^3}{18}$

$\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}$

$\text{Si}\left( n\pi \right) =-n\int_0^{\pi}{\cos nx\ln x\text{d}x}$

$\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}=\sum_{n=1}^{\infty}{\left( \int_0^{\pi}{\cos nx\ln x\text{d}x} \right) ^2}$

$\widetilde{f}\left( x \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n\cos nx}$

$\frac{a_{0}^{2}}{2}+\sum_{n=1}^{\infty}{a_{n}^{2}}=\frac{2}{\pi}\int_0^{\pi}{f^2\left( x \right) \text{d}x}=\frac{2}{\pi}\int_0^{\pi}{\ln ^2x\text{d}x}=4-4\ln \pi +2\ln ^2\pi$

$\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}=\sum_{n=1}^{\infty}{\left( \int_0^{\pi}{\cos nx\ln x\text{d}x} \right) ^2}=\frac{\pi^2}2$

$\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^5}}=\frac{269}{43200}\pi ^5,\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}\text{Si}\left( n\pi \right)}{n^5}}=\frac{4}{675}\pi ^5$

$\sum_{n=1}^{\infty}{\frac{\text{Si}(n\pi)}{n^5}}=\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n^4}\left( \pi \ln \pi -\pi \right)}-\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}$

\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\text{d}\left( \frac{1}{2}x^2\ln x-\frac{3}{4}x^2 \right)}}\\&=\sum_{n=1}^{\infty}\frac1{n^2}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \cos nx\text{d}x}}\\&=\sum_{n=1}^{\infty}{\frac{1}{n^2}\int_0^{\pi}{\cos nx\text{d}\left( \frac{11}{36}x^3-\frac{1}{6}x^3\ln x \right)}}\\&=\left( \frac{11}{36}\pi ^3-\frac{\pi ^3}{6}\ln \pi \right) \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n^2}}+\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}\\&=-\left( \frac{11}{36}\pi ^3-\frac{\pi ^3}{6}\ln \pi \right) \frac{\pi ^2}{12}+\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}.\end{align*}
\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\text{d}\left( \frac{25}{288}x^4-\frac{1}{24}x^4\ln x \right)}}\\&=\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{25}{288}x^4-\frac{1}{24}x^4\ln x \right) \cos nx\text{d}x}}.\end{align*}

$\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^5}}=\frac{7\pi ^4}{720}\left( \pi \ln \pi -\pi \right) +\frac{\pi ^2}{12}\left( \frac{11\pi ^3}{36}-\frac{\pi ^3}{6}\ln \pi \right) -\frac{\pi}{2}\cdot \frac{\pi ^4\left( 137-60\ln \pi \right)}{7200}=\frac{269}{43200}\pi ^5$

$\sum_{n=1}^{\infty}{\frac{\text{Si}^2\left( n\pi \right)}{n^4}}=\frac{\pi ^4}{27}$

$\frac{\text{Si}\left( n\pi \right)}{n^2}=\frac{\left( -1 \right) ^{n-1}}{n}\left( \pi \ln \pi -\pi \right) -\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}$

\begin{align*}\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^2}\sin nx}&=\left( \pi \ln \pi -\pi \right) \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n}\sin nx}-\sum_{n=1}^{\infty}{\sin nx\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}\\&=\frac{\left( \pi \ln \pi -\pi \right) x}{2}-\frac{\pi \left( x\ln x-x \right)}{2}.\end{align*}

$\sum_{n=1}^{\infty}{\frac{\text{Si}^2\left( n\pi \right)}{n^4}}=\frac{2}{\pi}\int_0^{\pi}{\left( \frac{\left( \pi \ln \pi -\pi \right) x}{2}-\frac{\pi \left( x\ln x-x \right)}{2} \right) ^2\text{d}x}=\frac{\pi ^4}{27}$

\begin{align*}\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^3}}&=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\frac{\sin nx}{x}\text{d}x}}=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\text{d}x}\int_0^{+\infty}{\text{e}^{-xy}\text{d}y}}\\&=\int_0^{+\infty}{\text{d}y}\int_0^{\pi}{\text{e}^{-xy}}\frac{x^3-3\pi x^2+2\pi ^2x}{12}\text{d}y=\frac{5}{72}\pi ^3.\end{align*}

## 美国数学月刊无穷乘积与西西新年祝福题

$$\sum_{i=1}^{\infty}{a_{i}^{-6}\left(\frac{\sin a_i-\sinh a_i}{\cos a_i+\cosh a_i}\right)^2}=\frac{1}{80}.$$

1、求无穷积分$$\int_{1/2}^{+\infty}{\int_{1/2}^{+\infty}{\int_{1/2}^{+\infty}{\frac{dxdydz}{\prod\limits_{cyc}{\left[242x^5-\left(y-1\right)^5-\left(z+1\right)^5\right]}}}}}.$$

2、设${\left( {1 + \frac{1}{x}} \right)^x} = e\left( {1 - \sum\limits_{k = 1}^\infty {\frac{{{d_k}}}{{{{\left( {\frac{{11}}{{12}} + x} \right)}^k}}}} } \right),$求
${\sum\limits_{k = 1}^\infty {\frac{1}{{1 + d_k^2}}} }.$

(2017年10月AMM征解题)求证

$\prod\limits_{j \ge 1} {{e^{ - 1/j}}\left( {1 + \frac{1}{j} + \frac{1}{{2{j^2}}}} \right)} = \frac{{{e^{\pi /2}} + {e^{ - \pi /2}}}}{{\pi {e^\gamma }}}.$

${x_n} = \prod\limits_{k = 1}^n {\left( {1 + \frac{1}{k} + \frac{1}{{2{k^2}}}} \right)} = \prod\limits_{k = 1}^n {\frac{{{{\left( {2k + 1} \right)}^2} + 1}}{{{{\left( {2k} \right)}^2}}}} ,$
\begin{align*}\frac{{\prod\limits_{k = 1}^{2n} {\left( {1 + \frac{1}{{{k^2}}}} \right)} }}{{{x_n}}} &= \frac{{\prod\limits_{k = 1}^{2n} {\frac{{{k^2} + 1}}{{{k^2}}}} }}{{\prod\limits_{k = 1}^n {\frac{{{{\left( {2k + 1} \right)}^2} + 1}}{{{{\left( {2k} \right)}^2}}}} }} = \frac{{\left( {{1^2} + 1} \right)\left( {{2^2} + 1} \right)\left( {{4^2} + 1} \right) \cdots \left[ {{{\left( {2n} \right)}^2} + 1} \right]}}{{{1^2}{3^2} \cdots {{\left( {2n - 1} \right)}^2}\left[ {{{\left( {2n + 1} \right)}^2} + 1} \right]}}\\&= 2\prod\limits_{k = 1}^n {\left( {1 + \frac{1}{{4{k^2}}}} \right) \cdot } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{{4{n^2} + 4n + 2}},\end{align*}

$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{{2n + 1}} = \frac{\pi }{2}.$

$\frac{{\sinh \left( {\pi x} \right)}}{{\pi x}} = \prod\limits_{k = 1}^\infty {\left( {1 + \frac{{{x^2}}}{{{k^2}}}} \right)}$

$\prod\limits_{k = 1}^\infty {\left( {1 + \frac{1}{{{k^2}}}} \right)} = \frac{{{e^\pi } - {e^{ - \pi }}}}{{2\pi }},\quad \prod\limits_{k = 1}^\infty {\left( {1 + \frac{1}{{4{k^2}}}} \right)} = \frac{{{e^{\pi /2}} - {e^{ - \pi /2}}}}{\pi },$

${H_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} = \ln n + \gamma + o\left( 1 \right),$
$\mathop {\lim }\limits_{n \to \infty } n\prod\limits_{k = 1}^n {{e^{ - 1/k}}} = {e^{ - \gamma }}.$

$\frac{{\frac{{{e^\pi } - {e^{ - \pi }}}}{{2\pi }}}}{{{e^\gamma } \times \frac{\pi }{2} \times \frac{{{e^{\pi /2}} - {e^{ - \pi /2}}}}{\pi }}} = \frac{{{e^{\pi /2}} + {e^{ - \pi /2}}}}{{\pi {e^\gamma }}}.$

\begin{align*}\cosh \left( {\pi x} \right) &= \prod\limits_{n = 1}^\infty {\left( {1 + \frac{{4{x^2}}}{{{{\left( {2n - 1} \right)}^2}}}} \right)} ,&\frac{{\sinh \left( {\pi x} \right)}}{{\pi x}} &= \prod\limits_{n = 1}^\infty {\left( {1 + \frac{{{x^2}}}{{{n^2}}}} \right)} ,\\\cos \left( {\pi x} \right) &= \prod\limits_{n = 1}^\infty {\left( {1 - \frac{{4{x^2}}}{{{{\left( {2n - 1} \right)}^2}}}} \right)} ,&\frac{{\sin \left( {\pi x} \right)}}{{\pi x}} &= \prod\limits_{n = 1}^\infty {\left( {1 - \frac{{{x^2}}}{{{n^2}}}} \right)} ,\end{align*}

\begin{align*}\sqrt 2 \sin \left( {\frac{{x + 1}}{4}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 + \frac{{{{\left( { - 1} \right)}^n}x}}{{2n + 1}}} \right)} ,\\\sqrt {x + 1} \sin \left( {\frac{{\sqrt {x + 1} }}{2}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 - \frac{x}{{4{n^2} - 1}}} \right)} ,\\- \sqrt {x - 1} \mathrm{csch}\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{{\sqrt {x - 1} }}{2}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 - \frac{x}{{4{n^2} + 1}}} \right)} ,\\- \sqrt { - x - 1} \mathrm{csch}\left( {\frac{\pi }{{\sqrt a }}} \right)\sin \left( {\frac{{\sqrt { - x - 1} }}{{\sqrt a }}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 + \frac{x}{{a{n^2} + 1}}} \right)} ,\\\frac{{{e^{ - \gamma x}}}}{{\Gamma \left( {1 + x} \right)}} &= \prod\limits_{n = 1}^\infty {\frac{{1 + x/n}}{{{e^{x/n}}}}},\end{align*}

\begin{align*}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} - {x^2}}}} &= \frac{1}{{2{x^2}}} - \frac{\pi }{{2x}}\cot \left( {\pi x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {{n^2} - {x^2}} \right)}^2}}}} &= - \frac{1}{{2{x^4}}} - \frac{{{\pi ^2}}}{{4{x^2}}}\mathrm{csc}^2\left( {\pi x} \right) + \frac{\pi }{{4{x^3}}}\cot \left( {\pi x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {2n - 1} \right)}^2} - {x^2}}}} &= \frac{\pi }{{4x}}\tan \left( {\frac{\pi }{2}x} \right),&& \left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left[ {{{\left( {2n - 1} \right)}^2} - {x^2}} \right]}^2}}}} &= \frac{{{\pi ^2}}}{{16{x^2}}}\sec \left( {\frac{\pi }{2}x} \right) - \frac{\pi }{{8{x^3}}}\tan \left( {\frac{\pi }{2}x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + {x^2}}}} &= \frac{\pi }{{2x}}\coth \left( {\pi x} \right) - \frac{1}{{2{x^2}}}, &&\left| x \right| < \infty\\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {2n - 1} \right)}^2} + {x^2}}}} &= \frac{\pi }{{4x}}\tanh \left( {\frac{\pi }{2}x} \right),&&\left| x \right| < \infty\end{align*}

The Weierstrass factorization theorem. Sometimes called the Weierstrass product/factor theorem.

Let $f$ be an entire function, and let $\{a_n\}$ be the non-zero zeros of $ƒ$ repeated according to multiplicity; suppose also that $ƒ''$ has a zero at $z= 0$ of order $m\geq 0$ (a zero of order $m=0$ at $z=0$ means $f(0)\neq 0$.
Then there exists an entire function $g$ and a sequence of integers $\{p_n\}$ such that

$f(z)=z^m e^{g(z)} \prod_{n=1}^\infty E_{p_n}\left(\frac{z}{a_n}\right).$

====Examples of factorization====

\begin{align*}\sin \pi z &= \pi z \prod_{n\neq 0} \left(1-\frac{z}{n}\right)e^{z/n} = \pi z\prod_{n=1}^\infty \left(1-\left(\frac{z}{n}\right)^2\right)\\\cos \pi z &= \prod_{q \in \mathbb{Z}, \, q \; \text{odd} } \left(1-\frac{2z}{q}\right)e^{2z/q} = \prod_{n=0}^\infty \left( 1 - \left(\frac{z}{n+\tfrac{1}{2}} \right)^2 \right) \end{align*}

The cosine identity can be seen as special case of
$\frac{1}{\Gamma(s-z)\Gamma(s+z)} = \frac{1}{\Gamma(s)^2}\prod_{n=0}^\infty \left( 1 - \left(\frac{z}{n+s} \right)^2 \right)$
for $s=\tfrac{1}{2}$.

Mittag-Leffler's theorem.

== Pole expansions of meromorphic functions ==
Here are some examples of pole expansions of meromorphic functions:

\begin{align*}\frac{1}{\sin(z)}&= \sum_{n \in \mathbb{Z}} \frac{(-1)^n}{z-n\pi}= \frac{1}{z} + 2z\sum_{n=1}^\infty (-1)^n \frac{1}{z^2 - (n\,\pi)^2},\\\cot(z) &\equiv \frac{\cos (z)}{\sin (z)}= \sum_{n \in \mathbb{Z}} \frac{1}{z-n\pi}= \frac{1}{z} + 2z\sum_{k=1}^\infty \frac{1}{z^2 - (k\,\pi)^2},\\\frac{1}{\sin^2(z)} &= \sum_{n \in \mathbb{Z}} \frac{1}{(z-n\,\pi)^2},\\\frac{1}{z \sin(z)}&= \frac{1}{z^2} + \sum_{n \neq 0} \frac{(-1)^n}{\pi n(z-\pi n)}= \frac{1}{z^2} + \sum_{n=1}^\infty \frac{(-1)^n}{n\,\pi} \frac{2z}{z^2 - (n\,\pi)^2}.\end{align*}

## 好题

(1) $\displaystyle I\left( a,b \right) =I\left( \frac{a+2b}{3},\sqrt[3]{b\frac{a^2+ab+b^2}{3}} \right)$.

(2) 数列$\{a_n\},\{b_n\}$满足$a_0=a,b_0=b$,且满足$a_{n+1}=\frac{a+2b}{3},b_{n+1}=\sqrt[3]{b_n\frac{a_n^2+a_nb_n+b_n^2}{3}}$,求证$\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\frac{I(1,1)}{I(a,b)}$.

Ramanujan's golden ratio equation
\begin{align*}R\left( e^{-2\pi} \right) &=\frac{e^{-\frac{2\pi}{5}}}{1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+\ddots}}}=\sqrt{\frac{5+\sqrt{5}}{2}}-\phi ,\\R\left( e^{-2\sqrt{5}\pi} \right) &=\frac{e^{-\frac{2\pi}{\sqrt{5}}}}{1+\frac{e^{-2\pi \sqrt{5}}}{1+\frac{e^{-4\pi \sqrt{5}}}{1+\ddots}}}=\frac{\sqrt{5}}{1+\left( 5^{3/4}\left( \phi -1 \right) ^{5/2}-1 \right) ^{1/5}}-\phi ,\end{align*}

\begin{align*}R(q) & = q^{\frac{1}{5}}\prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})} \\ &= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}.\end{align*}

Ramanujan–Sato series,https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series
$\frac{1}{\pi} = \frac{2 \sqrt 2}{99^2} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{26390k+1103}{396^{4k}}.$
Chudnovsky algorithm，https://en.wikipedia.org/wiki/Chudnovsky_algorithm
$\frac{1}{\pi} = 12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (545140134k + 13591409)}{(3k)!(k!)^3 \left(640320\right)^{3k + 3/2}}.$此公式对$\pi$有非常好的计算性能.

Ramanujan's Hypergeometric Identity,http://mathworld.wolfram.com/RamanujansHypergeometricIdentity.html

$$\sum_{k=-\infty}^\infty 2^k = 0.$$

Plot the graphs of the functions $$f(x)=\dfrac{2(x^2+|x|-6)}{3(x^2+|x|+2)}+\sqrt{16-x^2}$$ and $$g(x)=\dfrac{2(x^2+|x|-6)}{3(x^2+|x|+2)}-\sqrt{16-x^2}$$ in $x\in[-4,4]$ on the same plane.

![enter image description here][1]

[1]: http://i.stack.imgur.com/9PdtB.jpg

$\sum_{n=1}^{\infty} \frac{n^{13}}{e^{2\pi n} - 1} = \frac{1}{24}.$

sinx小于x小于tanx，Young不等式，等周问题

https://math.stackexchange.com/a/411763/165013

https://math.stackexchange.com/a/842310/165013

https://math.stackexchange.com/a/2323155/165013

https://math.stackexchange.com/a/718750/165013

https://math.stackexchange.com/a/83952/165013

https://math.stackexchange.com/a/61727/165013

## 曲面积分计算

(2012年中科院考研题)设$\rho (x,y,z)$是原点$O$到椭球面$\frac{x^2}2+\frac{y^2}2+z^2=1$的上半部分(即满足$z\geq 0$的部分) $\Sigma$的任一点$(x,y,z)$处的切面的距离,求积分$\iint_\Sigma \frac z{\rho (x,y,z)}dS.$

$I=\iint_{\Sigma}{\frac{z}{\rho \left( x,y,z \right)}dS}=\frac{1}{2}\iint_{\Sigma}{z\sqrt{x^2+y^2+z^2}dS}.$

\begin{align*}dS&=\sqrt{1+\left(\frac{\partial \varphi}{\partial x}\right)^2+\left(\frac{\partial \varphi}{\partial x}\right)^2}dxdy=\sqrt{1+\left(\frac{-x}{2z}\right)^2+\left(\frac{-y}{2z}\right)^2}dxdy\\&=\frac1{2z}\sqrt{x^2+y^2+4z^2}dxdy.\end{align*}

\begin{align*}I&=\iint_{\Sigma}{\frac{z}{\rho \left( x,y,z \right)}dS}=\frac{1}{2}\iint_{\Sigma}{z\sqrt{x^2+y^2+z^2}dS}\\&=\frac{1}{4}\iint_D\sqrt{x^2+y^2+z^2}\sqrt{x^2+y^2+4z^2}dxdy=\frac{1}{4}\iint_D\sqrt{1+\frac{x^2}2+\frac{y^2}2}\sqrt{4-x^2-y^2}dxdy\\&=\frac14\int_0^{2\pi}d\theta\int_0^{\sqrt{2}}r\sqrt{1+\frac{r^2}2}\sqrt{4-r^2}dr=\frac\pi4\int_0^{\sqrt{2}}\sqrt{1+\frac{u}2}\sqrt{4-u}du\\&=\frac\pi{4\sqrt{2}}\int_0^{\sqrt{2}}\sqrt{8+2u-u^2}du=\frac{\sqrt{2}\pi}{16}\left(\sqrt{10-6\sqrt{2}}+9\arcsin \frac{\sqrt 2-1}3+2\sqrt{2}+9\arcsin \frac13\right).\end{align*}

\begin{align*}&\int{\sqrt{8+2u-u^2}du}=u\sqrt{8+2u-u^2}-\int{\frac{u-u^2}{\sqrt{8+2u-u^2}}du}\\&=u\sqrt{8+2u-u^2}-\int{\frac{\left( 8+2u-u^2 \right) -8-u}{\sqrt{8+2u-u^2}}du}\\&=u\sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+\int{\frac{8+u}{\sqrt{8+2u-u^2}}du}\\&=\left( u-1 \right) \sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+\int{\frac{9}{\sqrt{9-\left( u-1 \right) ^2}}du}\\&=\left( u-1 \right) \sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+9\arcsin \frac{u-1}{3}+C,\end{align*}
$$\int{\sqrt{8+2u-u^2}du}=\frac{u-1}{2}\sqrt{8+2u-u^2}+\frac{9}{2}\arcsin \frac{u-1}{3}+C.$$

## 国外论坛好题

Sylvester equation https://en.wikipedia.org/wiki/Sylvester_equation
To seek the maximum value of $S=x_1x_2+x_2x_3+\cdots+x_nx_1$ on this domain:

$x_1+x_2+\cdots+x_n=0$ and
$x_1^2+x_2^2+\cdots+x_n^2=1$.

I have made some trivial observations:

1) $S\in[-1,1)$ by the rearrangement inequality.

2) We can make $S$ arbitrarily close to $1$ by increasing $n$.

3) An equivalent problem is to minimise $(x_1-x_2)^2+\cdots+(x_n-x_1)^2$.

But does the maximum have a meaningful closed form  for each $n$?
I propose to do a discrete Fourier transform. To this end put $\omega:=e^{2\pi i/n}$. In the following  all sums are over ${\mathbb Z}_n$, unless indicated otherwise. Let
$$y_k:={1\over n}\sum_l x_l\omega^{-kl}\ .$$
Since the $x_l$ are real we have $$y_{-k}=\overline{y_k}\tag{1}$$ for all $k$, furthermore $y_0=\sum_lx_l=0$. One has Parseval's formula
$$n\sum_k y_k\overline{ y_k}=\sum_l x_l^2=1\tag{2}$$
and the inversion formula
$$x_j=\sum_k y_k\omega^{jk}\ .\tag{3}$$
Using $(3)$ one computes
$$S=\sum_j x_jx_{j+1}=\ldots=n\sum_k|y_k|^2\omega^k\ .\tag{4}$$
At this point we have to distinguish the cases (a) $n=2m$, and (b) $n=2m+1$.

(a) If $n=2m$ then $\omega^m=-1$, and $(4)$ gives
\eqalign{S&=n\left(\sum_{k=1}^{m-1}|y_k|^2(\omega^k+\omega^{-k}) \ +|y_m|^2\omega^m\right)\cr &=n\left(\sum_{k=1}^{m-1}|y_k|^2\>2\cos{2k\pi\over n} \ -|y_m|^2\right)\ .\cr}
Given the conditions $(1)$ and $(2)$ it is easily seen that the optimal admissible choice of the $y_k$ is $$y_1=y_{-1}={1\over\sqrt{2n}}\>,\qquad y_k=0\quad(k\ne\pm1)\ .\tag{5}$$  This leads to
$$S_{\rm opt}=\cos{2\pi\over n}\ .$$
In particular when $n=4$ one obtains $S_{\rm opt}=0$, as indicated in Zubzub's answer.

(b) If $n=2m+1$ then $(4)$ gives
$$S=2n\sum_{k=1}^m |y_k|^2\cos{2k\pi\over n}\ ,$$
and the choice $(5)$ leads again to
$$S_{\rm opt}=\cos{2\pi\over n}\ .$$

In particular when $n=3$ one obtains $S_{\rm opt}=-{1\over2}$, and when $n=5$ one obtains $S_{\rm opt}=\cos{2\pi\over5}\doteq0.309$, as indicated in Zubzub's answer.

Let $$A=\left[ \begin{matrix} A_1& A_3\\ 0& A_2\\ \end{matrix} \right]\quad \text{and} \quad B=\left[ \begin{array}{c} B_1\\ B_2\\ \end{array} \right],$$and for any eigenvalue $s$ of $A$, we have
$$\mathrm{rank}\,\left[ A-sI_n,B \right] =n.$$
Prove there are a real matrix $K\in \mathbb{R}^{r\times n}$ and invertible matrix $T\in \mathbb{R}^{n\times n}$, such that
$$T\left( A+BK \right) T^{-1}=\left[ \begin{matrix} A_1& 0\\ 0& \bar{A}_2\\ \end{matrix} \right] ,\qquad TB=\left[ \begin{array}{c} \bar{B}_1\\ B_2\\ \end{array} \right],$$
where $A\in \mathbb{R}^{n\times n},B\in \mathbb{R}^{n\times r}$. Meanwhile,  $\bar{A}_2$ and $\bar{B}_1$ is real matrix of the proper dimension, $\bar{A}_2$ and $A_1$ Have eigenvalues that are not identical to each other.

This problem is about controllability of linear system and pole assignment, so I think we may try controllability canonical form, or let $K=(K_1,K_2)$, then determine the $K_1,K_2$ and $T$, but it seems so difficult.

The statement about the rank of $\left[A - s\,I, B\right]$ implies that the pair $(A,B)$ is [controllable](https://en.wikipedia.org/wiki/Hautus_lemma). Therefore the poles of the resulting matrix should be able to be placed anywhere. The similarity transformation should allow us to separate the eigenvalues/modes and therefore achieve the stated goal

$$T \left(A + B\,K\right) T^{-1} = \begin{bmatrix} A_1 & 0 \\ 0 & \bar{A}_2 \end{bmatrix}, \quad T\,B = \begin{bmatrix} \bar{B}_1 \\ B_2 \end{bmatrix},$$

given that

$$A = \begin{bmatrix} A_1 & A_3 \\ 0 & A_2 \end{bmatrix}, \quad B = \begin{bmatrix} B_1 \\ B_2 \end{bmatrix}.$$

For solving this it is easier to separate the problem in smaller problems. For this I will define $T$ and $K$ as

$$T = \begin{bmatrix} T_1 & T_2 \\ T_3 & T_4 \end{bmatrix}, \quad K = \begin{bmatrix} K_1 & K_2 \end{bmatrix}.$$

The last goal requires

$$T\,B = \begin{bmatrix} T_1\,B_1 + T_2\,B_2 \\ T_3\,B_1 + T_4\,B_2 \end{bmatrix} = \begin{bmatrix} \bar{B}_1 \\ B_2 \end{bmatrix}.$$

The bottom half of this goal might have infinitely many solution if there is any overlap in the span of $B_1$ and $B_2$. But this problem should be solvable in general, in which case $T_3=0$ and $T_4=I$ should always solve it. Using this then the inverse of $T$ can shown to be

$$T^{-1} = \begin{bmatrix} T_1 & T_2 \\ 0 & I \end{bmatrix}^{-1} = \begin{bmatrix} T_1^{-1} & -T_1^{-1}\,T_2 \\ 0 & I \end{bmatrix}.$$

Since nothing is specified about $\bar{B}_1$ then $T_1$ and $T_2$ could be anything for now as long as $T_1$ is invertible. The left hand side of the first goal can now be written as

$$T \left(A + B\,K\right) T^{-1} = \begin{bmatrix} T_1\,A_1\,T_1^{-1} + \bar{B}_1\,K_1\,T_1^{-1} & T_1\,A_3 + T_2\,A_2 - T_1\,A_1\,T_1^{-1}\,T_2 + \bar{B}_1\left(K_2 - K_1\,T_1^{-1}\,T_2\right) \\ B_2\,K_1\,T_1^{-1} & A_2 + B_2\left(K_2 - K_1\,T_1^{-1}\,T_2\right) \end{bmatrix}.$$

It can be shown that the top and bottom left half can be set equal to the goal by using $T_1=I$ and $K_1=0$. This allows the top and bottom right of the first goal equation to be simplified to

$$\begin{bmatrix} A_3 + T_2\,A_2 - A_1\,T_2 + \bar{B}_1\,K_2 \\ A_2 + B_2\,K_2 \end{bmatrix} = \begin{bmatrix} 0 \\ \bar{A}_2 \end{bmatrix}.$$

For the bottom half of this equation you could just use a pole placement algorithm to find $K_2$, such that none of the poles match those of $A_1$. This should be possible since the pair $(A_2,B_2)$ should be controllable. The top half can then be rewritten as

$$A_3 + T_2\,\bar{A}_2 - A_1\,T_2 + B_1\,K_2 = 0,$$

which can be transformed into a [Sylvester equation](https://en.wikipedia.org/wiki/Sylvester_equation)

$$A\,X + X\,B = C,$$

with $X = T_2$, $A = -A_1$, $B = \bar{A}_2$ and $C = -A_3 - B_1\,K_2$. This equation has an unique solution for $X$ when $A$ and $-B$ do not have a common eigenvalue. This is an identical constraint as mentioned by your problem statement.

So to solve this problem you can first do a pole placement with the pair $(A_2,B_2)$ to find $K_2$, avoiding the eigenvalues of $A_1$. And then solve a Sylvester equation after substituting in this obtained $K_2$ in order to find $T_2$. Using these values then the final solution can then be expressed using

$$T = \begin{bmatrix} I & T_2 \\ 0 & I \end{bmatrix}, \quad K = \begin{bmatrix} 0 & K_2 \end{bmatrix}.$$

## 一个偏微分方程求解

$$\frac{\partial J}{\partial t}=\frac{1}{4}\left( \frac{\partial J}{\partial x} \right) ^2-x^2-\frac{1}{2}x^4 \tag 1$$
$$\frac{\partial^2 J}{\partial t\partial x}=\frac{1}{2} \frac{\partial J}{\partial x}\frac{\partial^2 J}{\partial x^2} -2x-2x^3$$
Change of function :
$$\frac{\partial J}{\partial x}=u(x,t)\quad\to\quad \frac{\partial u}{\partial t} -\frac{1}{2} u\frac{\partial u}{\partial x}= -2x-2x^3 \tag 2$$
Characteristic system of equations :
$$\frac{dt}{1}=\frac{dx}{-\frac{1}{2} u}=\frac{du}{-2x-2x^3}$$
First family of characteristic curves, from $\quad -2\frac{dx}{ u}=\frac{du}{-2x-2x^3} :$
$$2udu-(8x+8x^3)dx=0 \quad\to\quad u^2-4x^2-2x^4=c_1$$
Second family of characteristic curves, from $\frac{dt}{1}=\frac{dx}{-\frac{1}{2} u} :$
$$dx+\frac{u}{2}dt=0=dx+\frac{\sqrt{c_1+4x^2+2x^4}}{2}dt$$
$$dt+\frac{2dx}{\sqrt{c_1+4x^2+2x^4}}=0 \quad\to\quad t+\int\frac{2dx}{\sqrt{c_1+4x^2+2x^4}}=c_2$$
The integral can be expressed on closed form. The formula involves a special function, namely the Elliptic Integral of the first kind :
http://www.wolframalpha.com/input/?i=integrate+2%2Fsqrt(C%2B4+x%5E2%2B2+x%5E4)&x=0&y=0

[![enter image description here][1]][1]

In interest of space and in order to make easier the writing, this big formula will be symbolized as :
$$\int\frac{2dx}{\sqrt{c_1+4x^2+2x^4}}=\Psi\left(c_1,x\right)$$
Where $\Psi$ is the above known function. Thus the second family of characteristic curves is :
$$t+\Psi\left(c_1,x\right)=c_2$$

The general solution of the PDE $(2)$ is expressed on the form of the implicit equation :
$$F\left(\left(u^2-4x^2-2x^4\right) \:,\: \left(t+\Psi\left(u^2-4x^2-2x^4\:,\:x\right)\right) \right)=0$$
where $F$ is any differentiable function of two variables.

The function $F$ might be determined according to a boundary condition which has to be derived from a given boundary condition of Eq.$(1)$. Nevertheless it appears doubtful to find a closed form for $F$ considering the complicated function $\Psi$.

Supposing that the function $F$ be determined, which is optimistic, a more difficult step comes after, to go from $u(x,t)$ to $J(x,y)$, which suppose possible to find a closed form for $\int u(x,t)dx$.

[1]: https://i.stack.imgur.com/DLyqp.jpg