自然数倒数平方和2

自然数倒数平方和2

写在前面

欧拉心算前十几项自然数倒数平方和，并注意到上一篇的神奇配凑后得出了 $ {\sum_{k=1}^{+\infin } \frac{1}{k^2} } = \frac{\pi ^2}{6} $ 的结果，但这并不严谨，尤其是涉及到无穷级数的收敛性．现在让我们利用夹逼定理，严格证明这个结论．

Solution

准备工作和预备知识：

\[\cot^2x < \frac{1}{x^2} < \csc^2x\hspace {0.3cm} ( 0 < x < \frac{\pi}{2} )\\ \cot ^2 x + 1 = \csc ^2 x\\ {\frac{(\cos x + i\sin x)^n}{(\sin x)^n}} = {\frac{(\cos nx + i\sin nx)}{(\sin x)^n}} = (\cot x + i)^n \]

使用二项式定理展开 \((\cot x + i)^n\)，得到

\[(\cot x + i)^n = [ \left (\begin{array}{cc} n \\ 0 \end{array} \right) (\cot x)^n - \left (\begin{array}{cc} n \\ 2 \end{array} \right) (\cot x)^{n-2}+...] + [ \left (\begin{array}{cc} n \\ 1 \end{array} \right) (\cot x)^{n-1}i - \left (\begin{array}{cc} n \\ 3 \end{array} \right) (\cot x)^{n-3}i+...] \]

令 $ n=2m+1 \hspace{0.3cm} (m \in \N ^* ) $ ， $ x_t = \frac{t\pi}{2m+1} \hspace{0.3cm} (t\in {x|1 \le x \le m ,x\in \N^* }) $，比对虚部，可得：

\[\begin{aligned} \frac{ \sin nx }{(\sin x)^n} &= [ \left (\begin{array}{cc} n \\ 1 \end{array} \right) (\cot x)^{n-1} - \left (\begin{array}{cc} n \\ 3 \end{array} \right) (\cot x)^{n-3}+...+(-1)^{(2m+1)} ] \\ &= [ \left (\begin{array}{cc} 2m+1 \\ 1 \end{array} \right) (\cot x)^{2m} - \left (\begin{array}{cc} 2m+1 \\ 3 \end{array} \right) (\cot x)^{2(m-1)}+... - (-1)^{m} ] \\ &= [ \left (\begin{array}{cc} 2m+1 \\ 1 \end{array} \right) (\cot^2 x)^{m} - \left (\begin{array}{cc} 2m+1 \\ 3 \end{array} \right) (\cot^2 x)^{(m-1)}+... - (-1)^{m} ] \end{aligned} \]

因为 $ \sin ({n x_t}) =\sin (t \pi ) =0 $，利用根与系数的关系，导出关系：

\[\begin{aligned} \sum_{k=1}^{m} { \cot^2 x_t } &= \frac{ \left ( \begin{array}{c} 2m+1 \\ 3 \end{array} \right ) } { \left ( \begin{array}{c} 2m+1 \\ 1 \end{array} \right ) } \\ &= \frac{2m(2m-1)}{6} \end{aligned} \]

于是有

\[\begin{aligned} \sum_{k=1}^{m} \cot^2{x_k} &< \sum_{k=1}^{m} \frac{1}{ {x_k}^2 } < \sum_{k=1}^{m} \csc^2x_k \\ \frac{2m(2m-1)}{6} &< \sum_{k=1}^{m}\frac{(2m+1)^2}{k^2\pi^2} < \frac{2m(2m-1)}{6} +m \\ \frac{2m(2m-1)\pi^2}{6(2m+1)^2} &< \sum_{k=1}^{m}\frac{1}{k^2} < \frac{\pi^2}{(2m+1)^2} (\frac{2m(2m-1)\pi^2}{6} +m ) \\ \end{aligned} \]

当 $ m \rightarrow \infty $ 时，根据夹逼定理

\[ \lim _{m \rightarrow \infty} \frac{2m(2m-1)\pi^2}{6(2m+1)^2} =\lim _{m \rightarrow \infty} \frac{\pi^2}{(2m+1)^2} (\frac{2m(2m-1)\pi^2}{6} +m ) = \frac{\pi^2}{6} \\ \lim _{m \rightarrow \infty} \sum_{k=1}^{m}\frac{1}{k^2}=\frac{\pi^2}{6} \]

我们就完成了一个相对初等的巴塞尔问题的证明。