几个恒等式证明 - Eufisky - The lost book
1-1+1-1+...=1/2?

几个恒等式证明

Eufisky posted @ 2018年9月16日 00:44 in 数学分析 with tags Ramanujan , 996 阅读

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\[\sum_{n=1}^N{\left( \left\{ x+\frac{n}{N} \right\} -\frac{1}{2} \right)}=\left\{ Nx \right\} -\frac{1}{2}.\]

若$a,b$是互质的正整数,证明
\[\int_0^1{\left( \left\{ ax \right\} -\frac{1}{2} \right) \left( \left\{ bx \right\} -\frac{1}{2} \right) dx}=\frac{1}{12ab}.\]
 
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*}
提示:考虑关于$f(x)$的微分方程.
 
5.证明Ramanujan的恒等式
\[\int_0^{\infty}{e^{-3\pi x^2}\frac{\sin\text{h}\pi x}{\sin\text{h}3\pi x}dx}=\frac{1}{\sqrt{3}e^{2\pi /3}}\sum_{n=0}^{\infty}{\frac{e^{-2n\left( n+1 \right) \pi}}{\left( 1+e^{-\pi} \right) ^2\left( 1+e^{-3\pi} \right) ^2\cdots \left( 1+e^{-\left( 2n+1 \right) \pi} \right) ^2}}.\]
参考:G. N. Watson 1936:The Final Problem : An Account of the Mock Theta Functions
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?

参考:这里


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 $(2\log 2) x + O(\log x)$.
 

 

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