# 巴塞尔问题

$\sum_{n=1}^\infin \frac{1}{n^2} = \lim_{n \to +\infty}\left(\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2}\right).$

## 欧拉对这个问题的研究

$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots.$

$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots.$

\begin{align} \frac{\sin x}{x} & {} = \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \\ & {} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots. \end{align}

$-\left(\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \cdots \right) = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.$

$-\frac{1}{6} = -\frac{1}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{n^2}.$

$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.$

## 黎曼ζ函数

$\zeta(s) = \sum_{n=1}^\infin \frac{1}{n^s}.$

s = 2，我们可以看出ζ(2)等于所有平方数的倒数之和：

$\zeta(2) = \sum_{n=1}^\infin \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6} \approx 1.644934.$

$\sum_{n=1}^N \frac{1}{n^2} < 1 + \sum_{n=2}^N \frac{1}{n(n-1)} = 1 + \sum_{n=2}^N \left( \frac{1}{n-1} - \frac{1}{n} \right) = 1 + 1 - \frac{1}{N} \; \stackrel{N \to \infty}{\longrightarrow} \; 2.$

$\zeta(2n)=\frac{(2\pi)^{2n}(-1)^{n+1}B_{2n}}{2\cdot(2n)!}$

## 严密的证明

$\sum_{k=1}^m \frac{1}{k^2} = \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{m^2}$

x为一个实数，满足0 < x < π/2，并设n为正整数。从棣莫弗定理和余切函数的定义，可得：

$\frac{\cos (nx) + i \sin (nx)}{(\sin x)^n} = \frac{(\cos x + i\sin x)^n}{(\sin x)^n} = \left(\frac{\cos x + i \sin x}{\sin x}\right)^n = (\cot x + i)^n.$

$(\cot x + i)^n = {n \choose 0} \cot^n x + {n \choose 1} (\cot^{n-1} x)i + \cdots + {n \choose {n-1}} (\cot x)i^{n-1} + {n \choose n} i^n$
$= \left[ {n \choose 0} \cot^n x - {n \choose 2} \cot^{n-2} x \pm \cdots \right] \; + \; i\left[ {n \choose 1} \cot^{n-1} x - {n \choose 3} \cot^{n-3} x \pm \cdots \right].$

$\frac{\sin (nx)}{(\sin x)^n} = \left[ {n \choose 1} \cot^{n-1} x - {n \choose 3} \cot^{n-3} x \pm \cdots \right].$

$0 = {{2m+1} \choose 1} \cot^{2m} x_r - {{2m+1} \choose 3} \cot^{2m-2} x_r \pm \cdots + (-1)^m{{2m+1} \choose {2m+1}}$

$p(t) := {{2m+1} \choose 1}t^m - {{2m+1} \choose 3}t^{m-1} \pm \cdots + (-1)^m{{2m+1} \choose {2m+1}}.$

$\cot ^2 x_1 + \cot ^2 x_2 + \cdots + \cot ^2 x_m = \frac{\binom{2m+1}3} {\binom{2m+1}1}= \frac{2m(2m-1)}6.$

$\csc ^2 x_1 + \csc ^2 x_2 + \cdots + \csc ^2 x_m =\frac{2m(2m-1)}6 + m = \frac{2m(2m+2)}6.$

$\frac{2m(2m-1)}6 < \left( \frac{2m+1}{\pi} \right) ^2 + \left( \frac{2m+1}{2 \pi} \right) ^2 + \cdots + \left( \frac{2m+1}{m \pi} \right) ^2 < \frac{2m(2m+2)}6.$

$\frac{\pi ^2}{6}\left(\frac{2m}{2m+1}\right)\left(\frac{2m-1}{2m+1}\right) < \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{m^2} < \frac{\pi ^2}{6}\left(\frac{2m}{2m+1}\right)\left(\frac{2m+2}{2m+1}\right).$

m趋于无穷大时，左面和右面的表达式都趋于π2/6，因此根据夹挤定理，有：

$\zeta(2) = \sum_{k=1}^\infin \frac{1}{k^2} = \lim_{m \to \infty}\left(\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{m^2}\right) = \frac{\pi ^2}{6}$

## 傅里叶级数的证明

$f(x) = \sum_{n=1}^{\infty}\frac{2(-1)^{n+1}}{n} \sin(nx)$

${\pi^2 \over 3} = {1 \over 2\pi} \int_{-\pi}^{\pi} f^2(x) \, dx = \sum_{n=1}^{\infty}{1 \over 2\pi} \int_{-\pi}^{\pi} (2 \frac{(-1)^{n+1}}{n} \sin(nt) )^2 dt = 2 \sum_{n=1}^{\infty} {1 \over n^2}$

${\pi^2 \over 6} = \sum_{n=1}^{\infty} {1 \over n^2}$