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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TopNewestIf 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 Narayan Sharma · 2 months, 3 weeks ago

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– Soma Arjun · 2 months, 2 weeks ago

Wow. This is exactly what I was looking for. Thank you so much for such a wonderful reply.Log in to reply

When asking about "what have you tried", part of it is also to understand how much background you have, in order to provide a suitable reply. Your solution didn't give me great confidence in your understanding of calculus/anlysis, which is why I was unable to suggest taking a "high power" approach instead of a "crude summation like Riemann sum".

In future, by providing this context to others (e.g. I got this problem when working through Serge Lang's complex analysis book on this chapter), that will give you better replies. – Calvin Lin Staff · 2 months, 2 weeks ago

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What have you tried, where did you get stuck? – Calvin Lin Staff · 2 months, 3 weeks ago

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– Soma Arjun · 2 months, 3 weeks 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