# 巴塞尔问题的多种解法——怎么计算$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots$ ?

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

$\sum_{k=1}^{\infty}\frac{1}{(2k-1)^2}=\frac{\pi^2}{8}$

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

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

$x=n\cdot \pi,\mbox{ }(n = \pm1, \pm2, \pm3, \dots).$

\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 \notag\\
& {} = \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) \notag\cdots.
\end{align}
（PS：欧拉似乎没有证明这个无穷积，直到100年后魏尔斯特拉斯得到了他著名的“魏尔斯特拉斯分解定理”（Weierstrass factorization theorem，详情可见wiki相应条目）。利用这个方法得到函数时要特别小心，我以前看到的一个反例就可以说明这个问题：
http://tieba.baidu.com/p/1083636713 )

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

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

Lemma: 令$\omega_m = \frac{\pi}{2m+1}$,则
$\cot^2{\omega_m}+\cot^2{(2\omega_m)}+\cdots\cot^2{(m\omega_m)}=\frac{m(2m-1)}{3}.$

\begin{align*}
\sin{n\theta}&=\binom{n}{1}\sin{\theta}\cos^{n-1}{\theta}-\binom{n}{3}\sin^3{\theta}\cos^{n-3}{\theta}+\cdots \pm \sin^n{\theta}\\
&=\sin^n{\theta}\left(\binom{n}{1}\cot^{n-1}{\theta}-\binom{n}{3}\cot^{n-3}{\theta}+\cdots \pm 1\right)
\end{align*}

$\binom{n}{1}x^{m}-\binom{n}{3}x^{m-1}+\cdots \pm 1$

$\sum_{k=1}^{m}\cot^2{(k\omega_m)}<\sum_{k=1}^{m}\frac{1}{k^2\omega_m^2}<m+\sum_{k=1}^{m}\cot^2{(k\omega_m)}$

$\frac{m(2m-1)\pi^2}{3(2m+1)^2}<\sum_{k=1}^{m}\frac{1}{k^2}<\frac{m(2m-1)\pi^2}{3(2m+1)^2}+\frac{m\pi^2}{(2m+1)^2}$

$\frac{1}{n^2}=\int_{0}^1\int_0^1 x^{n-1}y^{n-1}dxdy$

$\sum_{n=1}^{\infty}\frac{1}{n^2}=\int_{0}^1\int_0^1\left(\sum_{n=1}^{\infty}(xy)^{n-1}\right)dxdy=\int_{0}^1\int_0^1 \frac{1}{1-xy}dxdy$

$\sum_{n=1}^{\infty}\frac{1}{n^2}=2\iint_S\frac{1}{1-u^2+v^2}dudv$
$S$是由点$(0,0),(1/2,-1/2),(1,0),(1/2,1/2)$构成的正方形，利用正方形的对称性，那么
\begin{align*}
2\iint_S\frac{1}{1-u^2+v^2}dudv&=4\int_{0}^{1/2}\int_{0}^{u}\frac{1}{1-u^2+v^2}dvdu+4\int_{1/2}^{1}\int_{0}^{1-u}\frac{1}{1-u^2+v^2}dvdu\\
\end{align*}

\begin{align*}\sum_{n=1}^{\infty}\frac{1}{n^2}&=4\int_0^{1/2}\frac{\arcsin{u}}{\sqrt{1-u^2}}du+4\int_{1/2}^{1}\frac{1}{\sqrt{1-u^2}}\left(\frac{\pi}{4}-\frac{\arcsin{u}}{2}\right)du\\
&=[2\arcsin{u}^2]_0^{1/2}+[\pi\arcsin{u}-\arcsin{u}^2]_{1/2}^{1}\\
&=\frac{\pi^2}{18}+\frac{\pi^2}{2}-\frac{\pi^2}{4}-\frac{\pi^2}{6}+\frac{\pi^2}{36}\\
&=\frac{\pi^2}{6}
\end{align*}

（未完待续..）

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