Prove that [latex]\sum\limits_{n=-\infty}^{\infty} \frac{(-1)^{n}}{(n+a)^{2}}=\frac{\pi^{2} \cos \pi a}{\sin ^{2} \pi a}[/latex] where [latex]a[/latex] is real and different from [latex]0, \pm 1, \pm 2, \ldots[/latex]

Let [latex]f(z)=1 /(z+a)^{2}[/latex] which has a double pole at [latex]z=-a[/latex]. Residue of [latex]\pi \csc \pi z /(z+a)^{2}[/latex] at [latex]z=-a[/latex] is [latex] \underset{z \rightarrow-a}{\lim} \frac{d}{d z}\left\{(z+a)^{2} \cdot \frac{\pi \csc \pi z}{(z+a)^{2}}\right\}=-\pi^{2} \csc \pi a \cot \pi a [/latex] Then, by Problem 7.29, [latex]\sum\limits_{n=-\infty}^{\infty} \frac{(-1)^{n}}{(n+a)^{2}}=-(\text { sum of residues })=\pi^{2} \csc \pi a \cot \pi a=\frac{\pi^{2} \cos \pi a}{\sin ^{2} \pi a}[/latex]

Question 7.30: Prove that ∑n=-∞^∞ (-1)^n/(n + a)² = π² cos πa/sin² πa where...

Schaum’s Outline of Complex Variables

Prove that $\sum\limits_{n=-\infty}^{\infty} \frac{(-1)^{n}}{(n+a)^{2}}=\frac{\pi^{2} \cos \pi a}{\sin ^{2} \pi a}$ where $a$ is real and different from $0, \pm 1, \pm 2, \ldots$

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