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I want to calculate $$\sum_{n=-\infty}^{n=\infty}\frac{1}{n^2+n+1}$$ The hint given is to calculate the integral $$\oint \frac{\pi\cot{\pi z}}{z^2+z+1}dz$$ over the square with corners $\pm(N+1/2)(1\pm i)$ where $N\in \mathbb{N}$ and then let $N\to\infty$.

My attempt was to parametrize the integral with and then integrate $\int_0^{2N+1}\frac{\pi\cot{\pi(N+1/2+i(N+1/2-t))}}{(N+1/2+i(N+1/2-t))^2+(N+1/2+i(N+1/2-t))+1}(-i)\;dt$ etc. but this seems far too complicated to be the intended solution.

stack_math
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    you need to use the residue theorem – Masacroso Nov 22 '22 at 12:45
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    The whole point of using this contour integral is that the residue theorem makes it easy. The poles of $\pi\cot(\pi z)$ are exactly the integers and it has residue $1$ at each of them. So what are the poles of $\frac{\pi\cot{\pi z}}{z^2+z+1}$ and what are its residues? The hardest part of the calculation is factoring $z^2+z+1$. – GEdgar Nov 22 '22 at 13:39

3 Answers3

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Without complex analysis.

If the goal is to compute $$\sum_{n=-\infty}^{n=+\infty}\frac{1}{n^2+n+1}$$ start with the partial sums $$S_p=\sum_{n=-p}^{n=+p}\frac{1}{n^2+n+1}=\sum_{n=-p}^{n=+p}\frac{1}{(n-a)(n-b)}=\frac 1{a-b}\sum_{n=-p}^{n=+p}\left(\frac 1{n-a}-\frac 1{n-b}\right)$$ where $$a=-\frac{1+i \sqrt{3}}{2} \qquad \qquad b=-\frac{1-i \sqrt{3}}{2}$$ Now, using $$\sum_{n=-p}^{n=+p}\frac{1}{n-c}=\frac{1}{c}+\psi ^{(0)}(p+1-c)-\psi ^{(0)}(p+1+c)+-\psi ^{(0)}(-c)+\psi ^{(0)}(c)$$ Using it twice, using series and converting the complex numbers $$S_p=\frac{2 \pi }{\sqrt{3}}\tanh \left(\frac{\pi\sqrt{3} }{2}\right)-\frac 2p+\frac 1{p^2}+O\left(\frac{1}{p^3}\right)$$

Claude Leibovici
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Note that the only poles of your functions are $z_1=-\frac{1}{2}-\frac{i\sqrt{3}}{2}$ and $z_2=-\frac{1}{2}+\frac{i\sqrt{3}}{2}$, and then you need to compute the residues at those poles. For example, for $z_1=-\frac{1}{2}-\frac{i\sqrt{3}}{2}$, we have $$Res\left(f,-\frac{1}{2}-\frac{i\sqrt{3}}{2}\right) = \lim_{z \rightarrow -\frac{1}{2}-\frac{i\sqrt{3}}{2}} \frac{\left(z-(-\frac{1}{2}-\frac{i\sqrt{3}}{2})\right)\pi\cot(\pi z)}{\left(z-(-\frac{1}{2}-\frac{i\sqrt{3}}{2})\right)\left(z-(-\frac{1}{2}+\frac{i\sqrt{3}}{2})\right)} \\[3em] = -\frac{\pi\cot(-\frac{\pi}{2}-\frac{i\pi\sqrt{3}}{2})}{i\sqrt{3}} = -\frac{i\pi\tanh\left(\frac{\pi\sqrt{3}}{2}\right)}{i\sqrt{3}} = -\frac{\pi\tanh\left(\frac{\pi\sqrt{3}}{2}\right)}{\sqrt{3}}.$$ Similarly for the other pole, and then $\sum_{n=-\infty}^{\infty} \frac{1}{n^2+n+1} = -\sum Res(f,z_k)$.

Mathick
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Noting $$ n^2+n+1=(n+\frac12)^2+(\frac{\sqrt3}2)^2 $$ and using $$ \sum_{n=-\infty}^\infty\frac{1}{(n-a)^2+b^2} = \frac{\pi}{b} \frac{\sinh(2\pi b)}{\cosh(2\pi b)-\cos(2\pi a)} $$ from here, one has $$ \sum_{n=-\infty}^\infty\frac{1}{n^2+n+1} = \frac{\pi}{b} \frac{\sinh(2\pi b)}{\cosh(2\pi b)-\cos(2\pi a)}\bigg|_{a=-\frac12,b=\frac{\sqrt3}2}=\frac{2 \pi }{\sqrt{3}}\tanh \left(\frac{\pi\sqrt{3} }{2}\right). $$

xpaul
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