证明 $$\sum\limits_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}} = \ln 3 - \frac{\pi }{4}.$$

解.(r9m)首先有

$$\begin{align*}S&=\sum\limits_{n=1}^{\infty} \arctan \frac{10n}{(3n^2+2)(9n^2-1)} \\&= \sum\limits_{n=1}^{\infty} \arg \left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4\left(1+\frac{2}{3n^2}\right)\left(1-\frac{1}{9n^2}\right)}\right).\end{align*}$$

分母里的无穷乘积$\displaystyle \prod\limits_{n=1}^{\infty}\left(1+\frac{2}{3n^2}\right)$和$\displaystyle \prod\limits_{n=1}^{\infty}\left(1-\frac{1}{9n^2}\right)$ 可以被忽略,当它们收敛于实数时.

因此

$$\begin{align*}S&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4}\right).\end{align*}$$

对分子进行分解

$$(3n^2+2)(9n^2-1)+10in = (n-i)(3n+i)(3n+i+1)(3n+i-1).$$

我们得

$$S = \arg \prod\limits_{n=1}^{\infty}\frac{\left(1+\frac{i}{3n}\right)\left(1+\frac{i+1}{3n}\right)\left(1+\frac{i-1}{3n}\right)}{\left(1+\frac{i}{n}\right)}.$$

在$\displaystyle z = i,\frac{i}{3},\frac{i+1}{3},\frac{i-1}{3}$ 处运用$\displaystyle \frac{1}{\Gamma(z)} = ze^{\gamma z}\prod\limits_{n=1}^{\infty}\left(1+\frac{z}{n}\right)e^{-z/n}$ .

我们可以改写为

$$S = \arg \frac{-\Gamma(i)}{\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right)\Gamma\left(\frac{i-1}{3}\right)}.$$

另一方面运用Gauss-Legendre Triplication Formula

$$\Gamma(3z) = \frac{1}{2\pi}3^{3z - \frac{1}{2}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{3}\right)\Gamma\left(z+\frac{2}{3}\right).$$

令$z = \dfrac{i-1}{3}$我们有

$$\Gamma\left(\frac{i-1}{3}\right)\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right) = 2\pi 3^{-i+\frac{3}{2}}\Gamma(i-1)$$

因此 $$S = \arg \frac{-3^{i}\Gamma(i)}{\Gamma(i-1)} = \arg (3^{i}(1-i)) = \log 3 - \frac{\pi}{4}.$$

解法二.(robjohn)运用$\arctan(x)=\arg(1+ix)$,分解可知

$$\begin{align*}&1+\frac{10in}{\left(3n^2+2\right)\left( 9n^2-1\right)}\\=&\frac{\left(1-\frac in\right)\left(1+\frac i{3n-1}\right)\left(1+\frac i{3n+1}\right)\left(1+\frac i{3n}\right)}{1+\frac2{3n^2}}.\end{align*}$$

因此有

$$\begin{align*}&\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\=&\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right).\end{align*}$$

裂项可知

$$\begin{align*}&\sum_{n=1}^\infty\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\=&\lim_{m\to\infty}\sum_{n=1}^m\left[\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right)\right]\\=&-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\arctan\left(\frac1n\right)\\=&-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\left[\frac1n+O\left(\frac1{n^3}\right)\right]\\=&\log(3)-\frac\pi4.\end{align*}$$

练习题1.证明： $$\sum\limits_{n=1}^{\infty}\frac{1}{(n+1)\sqrt[p]{n}}\leq p,\,\,(p\ge1).$$

证:由Lagrange中值定理,我们有

$$\sqrt[p]{{n + 1}} - \sqrt[p]{n} = \frac{1}{p}{\xi ^{1/p - 1}} \ge \frac{1}{p}{\left( {n + 1} \right)^{1/p - 1}},\quad \xi  \in \left( {n,n + 1} \right).$$

因此 $$\frac{1}{{\left( {n + 1} \right)\sqrt[p]{n}}} = \frac{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}}{{\sqrt[p]{n} \cdot \sqrt[p]{{n + 1}}}} \cdot \frac{{{{\left( {n + 1} \right)}^{1/p - 1}}}}{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}} \le p\frac{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}}{{\sqrt[p]{n} \cdot \sqrt[p]{{n + 1}}}} = p\left( {\frac{1}{{\sqrt[p]{n}}} - \frac{1}{{\sqrt[p]{{n + 1}}}}} \right).$$

立即有

$$\sum\limits_{n = 1}^\infty  {\frac{1}{{\left( {n + 1} \right)\sqrt[p]{n}}}}  \le p\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{\sqrt[p]{n}}} - \frac{1}{{\sqrt[p]{{n + 1}}}}} \right)}  = p.$$

练习题2.设$\displaystyle S_n=\sum\limits_{k=1}^{n}a_k, p>1,c>1$,证明： $$\sum\limits_{n=1}^{\infty}\frac{S_n^p}{n^c}\le K\sum\limits_{n=1}^{\infty}\frac{(na_n)^p}{n^c},$$ 并求出$K$的最优值.

练习题3.设$a_n$是有界的正数列，$p>0$，证明：

证.注意到

$$\begin{align*}\sum\limits_{k = 1}^\infty  {{a_k}}  &= \sum\limits_{k = 1}^\infty  {\arctan \frac{1}{{{k^2} + k + 1}}}  = \sum\limits_{k = 1}^\infty  {\left( {\arctan \frac{1}{k} - \arctan \frac{1}{{k + 1}}} \right)}  = \frac{\pi }{4}\\\sum\limits_{k = 1}^\infty  {\frac{1}{{{k^4}}}}  &= \zeta \left( 4 \right) = \frac{{{\pi ^4}}}{{90}}.\end{align*}$$

由Cauchy-Schwarz不等式可知

$$\sum\limits_{k = 1}^N {\frac{1}{{{k^4}}}}  \cdot \sum\limits_{k = 1}^N {{a_k}}  \ge {\left( {\sum\limits_{k = 1}^N {\frac{{a_k^{1/2}}}{{{k^2}}}} } \right)^2}.$$

令$N\to\infty$,我们有 $$\sum\limits_{k = 1}^\infty  {\frac{{a_k^{1/2}}}{{{k^2}}}}  \le \sqrt {\frac{{{\pi ^4}}}{{90}} \cdot \frac{\pi }{4}}  = \sqrt {\frac{{{\pi ^5}}}{{360}}}  < \sqrt {\frac{\pi }{3}} .$$

也可通过放缩实现 $$\sum\limits_{k = 1}^\infty  {\frac{1}{{{k^4}}}}  = 1 + \sum\limits_{k = 2}^\infty  {\frac{1}{{{k^4}}}}  < 1 + \sum\limits_{k = 2}^\infty  {\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}}  = \frac{4}{3}.$$

练习题10.设$\displaystyle a_n > 0, S_n=\sum\limits_{k=1}^na_k$，证明：

(2)如法炮制.由柯西不等式我们得

$$\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} \sum\limits_{m = 1}^n {{a_m}} \ge {\left( {1 + 2 + \cdots + n} \right)^2} = \frac{1}{4}{n^2}{\left( {n + 1} \right)^2},$$

即

$$\frac{{2n + 1}}{{{a_1} + {a_2} + \cdots + {a_n}}} \le \frac{{4\left( {2n + 1} \right)}}{{{n^2}{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} .$$

因此

$$\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{{a_1} + {a_2} + \cdots + {a_n}}}} &\le 4\sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} } = 4\sum\limits_{m = 1}^\infty {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty {\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}}} } \\&= 4\sum\limits_{m = 1}^\infty {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty {\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right]} } = 4\sum\limits_{m = 1}^\infty {\frac{1}{{{a_m}}}}.\end{align*}$$

练习题11.设$\displaystyle a_n \ge 0, n=1,2,\ldots,\sum\limits_{n=1}^{\infty}a_n < \infty$,证明:

$$\sum\limits_{n=1}^{\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}} \le e \sum\limits_{n=1}^{\infty}a_n$$ ,且证明$e$是最优值.

此题再拓展下求证: $$\sum\limits_{n=1}^{\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}} \le e \sum\limits_{n=1}^{\infty}[1-\frac{1}{2(n+1)}]a_n.$$

练习题12.如果正项级数$\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{p_n}$收敛,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}\frac{n^2}{(p_1+p_2+\cdots+p_n)^2}p_n$也收敛.

练习题13.设$\displaystyle \sum\limits_{n=1}^{\infty}a_n$为正项级数,且$\displaystyle \sum\limits_{k=1}^{n}(a_k-a_n)$对$n$有界,$a_n$单调递减趋于$0$,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}a_n$收敛.

练习题14.设级数$\displaystyle \sum\limits_{n=1}^{\infty}a_n$收敛, $\displaystyle \sum\limits_{n=1}^{\infty}(b_{n+1}-b_n)$绝对收敛,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}a_nb_n$收敛.

练习题15.设$a_n>0,\left\{ a_n-a_{n+1}\right\}$为一个严格递减的数列.如果$\sum_{n=1}^{\infty}a_n$收敛。试证： $$\lim\limits_{n \to \infty}\left( \cfrac{1}{a_{n+1}}-\cfrac{1}{a_n}\right)=+\infty.$$

练习题16.能否构造一个收敛数列$\sum\limits_{n=1}^{\infty}a_n$,使得级数$\sum\limits_{n=1}^{\infty}a_n^3$发散.

练习题17.设$\lim \limits_{n\rightarrow +\infty}x_n=+\infty$,正项级数$\sum\limits_{n=1}^{\infty}y_n$收敛,设$n_0$是某一自然数,

练习题18.设$\sum\limits_{n=1}^{\infty}a_n$是一正项收敛级数,且有$a_{n+1}< \frac{1}{2}(a_n+a_{n+2}),\,\frac{1}{a_{n+2}}-\frac{1}{a_{n+1}}\le \frac{1}{3}(\frac{1}{a_{n+3}}-\frac{1}{a_{n}})$,

1.证明: $\sum_{k=1}^n\sin k^2$无界.

参看: http://www.zhihu.com/question/29094450

2.证明 $$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\sin \sqrt k }  = 0.$$

Proof.What you require here are the Eisenstein series. In particular the evaluation of

本文主要展示并分享有关Euler Sum的若干求解问题。

问题一：求 $$\sum_{n=1}^\infty{\frac{H_n}{n^q}}$$

Solution.

参考文献

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