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 Post subject: Calculate the series
PostPosted: Mon Jul 11, 2016 9:16 am 
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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 series
PostPosted: Mon Jul 11, 2016 9:17 am 
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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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