calculus - Evaluating sums using residues $(-1)^n/n^2$

I am an alien towards compelx analysis, with very little know I am posing a question, who someone may want to help with.

Evaluate:

$$\frac{1}{4}\cdot \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$$

In disguise this is similar to $\zeta(2)$ but how can this be done using residues, and complex analysis?

I need some help. I am just interested.

The answer is $\displaystyle \frac{\pi^2}{48}$

Answer

Using $\boldsymbol{\pi\csc(\pi z)}$

Since $\pi\csc(\pi z)$ has residue $(-1)^n$ at $z=n$ for $n\in\mathbb{Z}$, we will use the contours
$$\gamma_\infty=\lim\limits_{R\to\infty}Re^{2\pi i[0,1]}\qquad\text{and}\qquad\gamma_0=\lim\limits_{R\to0}Re^{2\pi i[0,1]}$$
To sum over all $n\in\mathbb{Z}$ except $n=0$, we use the difference of the contours, which circles the non-zero integers once counter-clockwise.
$$\begin{align} 2\sum_{n=1}^\infty\frac{(-1)^n}{n^2} &=\frac1{2\pi i}\left(\int_{\gamma_\infty}\frac{\pi\csc(\pi z)}{z^2}\mathrm{d}z-\int_{\gamma_0}\frac{\pi\csc(\pi z)}{z^2}\mathrm{d}z\right)\\ &=\color{#C00000}{\frac1{2\pi i}\int_{\gamma_\infty}\frac{\pi\csc(\pi z)}{z^2}\mathrm{d}z}-\operatorname*{Res}_{z=0}\left(\color{#00A000}{\frac{\pi\csc(\pi z)}{z^2}}\right)\\ &=\color{#C00000}{0}-\operatorname*{Res}_{z=0}\left(\color{#00A000}{\frac1{z^2}\frac\pi{\pi z-\pi^3z^3/6+O\left(z^5\right)}}\right)\\ &=\color{#C00000}{0}-\operatorname*{Res}_{z=0}\left(\color{#00A000}{\frac1{z^3}+\frac{\pi^2}{6z}+O(z)}\right)\\ &=-\frac{\pi^2}6 \end{align}$$
because, for $k\in\mathbb{Z}$ and $|z|=\pi\left(k+\frac12\right)$, $|\csc(z)|\le1$.

Therefore,
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=-\frac{\pi^2}{12}$$

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Extending A Previous Result

In this answer, it is shown that
$$\sum_{k=1}^\infty\frac1{k^2}=\frac{\pi^2}6$$
Note that
$$\begin{align} \hphantom{=}&\frac1{1^2}{+}\frac1{2^2}+\frac1{3^2}{+}\frac1{4^2}+\frac1{5^2}{+}\frac1{6^2}+\frac1{7^2}+\dots\\ \hphantom{=}&\hphantom{\frac1{1^2}}\color{#C00000}{-\frac2{2^2}\hphantom{+\frac1{3^2}}-\frac2{4^2}\hphantom{+\frac1{5^2}}-\frac2{6^2}\hphantom{+\frac1{7^2}}-\dots}\\ =&\frac1{1^2}{-}\frac1{2^2}+\frac1{3^2}{-}\frac1{4^2}+\frac1{5^2}{-}\frac1{6^2}+\frac1{7^2}-\dots \end{align}$$
where the series in red is two times one quarter of the series above it; that is, one half of the series above it. Therefore, the alternating series is one half of the non-alternating series; that is,

$$\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}$$