Prove that

\[ \displaystyle \sum_{-\infty}^{\infty} \dfrac1{n+a} = \pi \cot (\pi a ) . \]

Prove that

\[ \displaystyle \sum_{-\infty}^{\infty} \dfrac1{n+a} = \pi \cot (\pi a ) . \]

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TopNewestWhat have you tried, where did you get stuck? – Calvin Lin Staff · 3 days, 2 hours ago

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– Soma Arjun · 2 days, 14 hours ago

Thank you so much for the response. I don't know how to deal with it. I know about General Harmonic series. But i am not getting any idea about how to solve a summation from "negative infinity to positive infinity" which also involves a constant "a" .Log in to reply

If you ask me how to do this by complex analysis , there is a much simpler answer.

i am using the summation theorem, first take a look here if necessary to see how all this happens. Let the function be \(\displaystyle f(z)=\frac{1}{z+a}\) . It has a simple pole at \(z=-a\) and by summation theroem we need to evaluate the residue of \(\pi\cot (\pi z)f(z)\) at \(z=-a\)

The residue is therefore \(\displaystyle R=\lim_{z\to -a}(z+a) \frac{\pi\cot\pi z}{z+a}=-\pi\cot\pi a\)

The answer is therefore the negative of the residue ,

\(\displaystyle \sum_{-\infty}^\infty \frac{1}{n+a}=\pi\cot \pi a\)

The series is actually the representation of \(\psi(a)\) where \(\psi(.)\) is the Digamma Function. – Aditya Sharma · 1 day, 8 hours ago

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