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 Post subject: Calculate the seriesPosted: Mon Jul 11, 2016 9:16 am

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Evaluate the series:

$$\mathbf{(a)} \; \sum_{n=-\infty}^{\infty}\frac{1}{n^2+4} \quad \text{and} \quad \mathbf{(b)} \; \sum_{n=-\infty}^{\infty}(-1)^n \frac{1}{n^2+4}$$

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 Post subject: Re: Calculate the seriesPosted: Mon Jul 11, 2016 9:17 am

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
I'll do the first one and give a hint for the second one.

*******************************************************
I'm proving the more general: $\displaystyle \sum_{n=-\infty}^{\infty}\frac{1}{n^2+a^2}=\frac{\pi}{a}\coth\left ( \pi a \right ) , \; a>0$

For this purpose I consider the function $\displaystyle g(z)=\pi\cot \left ( \pi z \right )f(z)=\frac{\pi \cot \left (\pi z \right )}{z^2+a^2}$ which has poles at $z_1=ai, \; z_2=-ai$. The residue at $z_1$ is $\displaystyle -\frac{\pi\coth(\pi a)}{2a}$ and similarly the residue at $z_2$ is equal to $\displaystyle -\frac{\pi\coth(\pi a)}{2a}$.

Hence:
$$\sum_{n=-\infty}^{\infty}\frac{1}{n^2+a^2}=-\sum_{residues}=\frac{\pi\coth(\pi a)}{a}$$

It is known that $\cot(ai)=-i\coth(a)$

For the second sum consider the function $\displaystyle g(z)=\pi \csc (\pi z)f(z)$ , evaluate the residues at all poles. The series will then be equal to the minus sum of the residues.

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