Bessel-like identity

By | May 16, 2015

Problem. Evaluate

    \[ \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{n^{2}}=\frac{\pi^{2}}{12}. \]

Proof. Let C_{N} denote the positively oriented boundary of the square whose edges lie along the lines

    \[ x=\pm\left(N+\frac{1}{2}\right)\pi\quad\text{and}\quad y=\pm\left(N+\frac{1}{2}\right)\pi, \]

where N is positive integer.
Define f\left(z\right)=\frac{1}{z^{2}\sin z}. Then at z=0, f has pole of order 3 because

    \[ \lim_{z\rightarrow0}\frac{z^{3}}{z^{2}\sin z}=1\neq0. \]

Note that

    \[ \frac{1}{z^{2}\sin z}=\frac{1}{z^{2}}\left(\frac{1}{z}+\frac{1}{3!}z+O\left(z^{3}\right)\right)=\frac{1}{z^{3}}+\frac{1}{6}\frac{1}{z}+O\left(z\right)\quad\left(0<\left|z\right|<\pi\right) \]

So

    \[ \mathrm{Res}\left(f,0\right)=\frac{1}{6}. \]

For k\in\mathbb{Z}\setminus\left\{ 0\right\}, f has simple pole at k\pi. So

    \[ \mathrm{Res}\left(f,k\pi\right)=\lim_{z\rightarrow k\pi}\left(z-k\pi\right)\frac{1}{z^{2}\sin z}=\frac{\left(-1\right)^{k}}{k^{2}\pi^{2}}. \]

In C_{N}, f has 2N+1 poles. So by Residue theorem, we have

    \[ \int_{C_{N}}f\left(z\right)dz=2\pi i\left(\frac{1}{6}+2\sum_{k=1}^{N}\frac{\left(-1\right)^{k}}{k^{2}\pi^{2}}\right). \]

Note that

    \[ \left|\sin z\right|\ge\left|\sin x\right|\quad\text{and}\quad\left|\sin z\right|\ge\left|\sinh y\right|. \]

Also note that on the vertical side of the square, \left|\sin z\right|\ge1 and \left|\sin z\right|>\sinh\left(\pi/2\right) on the horizontal sides. So there is a constant A such that \left|\sin z\right|\ge A for all z\in C_{N}.

So

    \begin{align*} \left|\int_{C_{N}}\frac{dz}{z^{2}\sin z}\right| & \le\frac{1}{A}\int_{C_{N}}\left|\frac{1}{z^{2}}\right|dz\\ & \le4\cdot\frac{2\left(N+\frac{1}{2}\right)\pi}{\left(N+\frac{1}{2}\right)^{2}\pi^{2}A}=\frac{16}{\left(2N+1\right)\pi A}. \end{align*}

So as N\rightarrow\infty,

    \[ \left|\int_{C_{N}}\frac{dz}{z^{2}\sin z}\right|\rightarrow0. \]

Therefore,

    \[ \frac{1}{6}+2\sum_{k=1}^{\infty}\frac{\left(-1\right)^{k}}{k^{2}\pi^{2}}=0, \]

i.e.,

    \[ \sum_{k=1}^{\infty}\frac{\left(-1\right)^{k+1}}{k^{2}}=\frac{\pi^{2}}{12}. \]


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