Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In an effort to find the infinite sum $\sum{\frac{1}{n^2+1}}$ I need to calculate the residues of $\frac{\pi \cot(\pi z)}{1+z^2}$ at poles $i, -i, 0$. I have two questions: first, how do I do this? I am used to denominators like $n^2$ but not $1 + n^2$, so it doesn't appear to be as simple as taking the Laurent expansion for $\cot(u)$ and finding the $u$ term. Second, why am I not also looking at the poles $1, 2, 3,\ldots$ since certainly $\sin(k\pi) = 0$ for $k \in \mathbb{N}$?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Factor ${z^2} + 1$ as ${z^2} + 1 = (z + i)(z - i)$. Then, using $i\cot (iz) = \coth (z)$, we have $${\text{Res}}(\pi \cot (\pi z);i) = \mathop {\lim }\limits_{z \to i} \frac{{(z - i)\pi \cot (\pi z)}}{{(z + i)(z - i)}} = - \frac{{\pi \coth (\pi )}}{2}$$ and $${\text{Res}}(\pi \cot (\pi z); - i) = \mathop {\lim }\limits_{z \to - i} \frac{{(z + i)\pi \cot (\pi z)}}{{(z + i)(z - i)}} = - \frac{{\pi \coth (\pi )}}{2}$$ so $$\sum\limits_{n = - \infty }^\infty {\frac{1}{{{n^2} + 1}} = } - \sum {{\text{Res}}(\pi \cot (\pi z);{z_k})} = \pi \coth (\pi ).$$ Therefore $$\sum\limits_{n = - \infty }^\infty {\frac{1}{{{n^2} + 1}} = } \sum\limits_{n = - \infty }^{ - 1} {\frac{1}{{{n^2} + 1}} + \sum\limits_{n = 0}^0 {\frac{1}{{{n^2} + 1}} + \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}} = } } } 1 + 2\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} = \pi \coth (\pi )$$ i.e. $$\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + 1}}} = \frac{{\pi \coth (\pi ) - 1}}{2}$$ because ${(- n)^2} = {n^2}$. Hope this helps.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.