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# Need proof for the following infinite series !!

Prove that

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

Note by Soma Arjun
8 months, 2 weeks ago

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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. · 8 months, 2 weeks ago

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Wow. This is exactly what I was looking for. Thank you so much for such a wonderful reply. · 8 months, 2 weeks ago

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(Assuming you know complex analysis,) do you see why complex analysis is the natural way to approach this problem?

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. Staff · 8 months, 2 weeks ago

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What have you tried, where did you get stuck? Staff · 8 months, 2 weeks ago

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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" . · 8 months, 2 weeks ago

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