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I'm trying to solve the sum: $$\sum_{n\in\mathbb{Z}}\frac{(-1)^n}{n^2}$$ I'm struggling with the $n=0$ term. Since $\frac{(-1)^n}{n^2}$ is even, can I just let this equal to $$2\sum_{n=1}^\infty\frac{(-1)^n}{n^2}?$$ This then converges. But I'm sure this is wrong, since I have basically ignored the $n=0$ term.

Any suggestions?

It might help if I provide some further comments:

In my supervisors's paper "The Berry-Keating operator on a lattice" they expand the discrete phase space translations in terms of a finite operator expansion (since the phase space is a torus) see appendix B. They then represent an operator $$Op_N(f)=\sum_{m,n\in\mathbb{Z}}f_{m,n}T^{m,n}=\sum_{m,n=0}^{N-1}\sum_{\mu,\nu\in\mathbb{Z}}f_{m+\mu N,n+\nu N}T^{m+\mu N,n+\nu N} =\sum_{m,n=0}^{N-1}g_{m,0}T^{m,n}\hspace{5mm}(1)$$ where $$T^{m+\mu N,n+\nu N}=(-1)^{m\nu +n\mu+\mu\nu N}T^{m,n}$$ and $$g_{m,n}=\sum_{\mu,\nu\in\mathbb{Z}}f_{m+\mu N,n+\nu N}(-1)^{m\nu +n\mu+\mu\nu N}.\hspace{3cm}(2)$$ The case which they are solving is $$Op(f)=\frac{\ell_\xi^2}{24}+\frac{\ell_\xi^2}{4\pi^2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(T^{n,0}+T^{-n,0}\right)\hspace{2cm}(3)$$ according to the paper $$g_{m,0}=\frac{\ell_\xi^2}{4\pi^2}\sum_{\mu\in\mathbb{Z}}\frac{(-1)^{m+\mu N}}{(m+\mu N)^2},\text{ and}$$ $$g_{0,0}=\frac{\ell_\xi^2}{24}+\frac{\ell_\xi^2}{2\pi^2}\sum_{\mu=1}^\infty\frac{(-1)^{\mu N}}{(\mu N)^2}.$$ I thought that they let $m=0$ in the $g_{m,0}$ term and used an even function argument to evaluate the sum, which is where my confusion came in. I now believe that they obtained the $g_{0,0}$ expression from $(3)$, but I don't really understand that since we have the expression for $g_{m,0}$ in terms of a sum over the integers $(2)$.

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    $\begingroup$ The expression $\frac{(-1)^0}{0^2}$ is undefined, so I suspect there is a mistake in the question. $\endgroup$ – Clive Newstead Jun 28 '17 at 10:52
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    $\begingroup$ The question should include $n\ne 0$ as expression is not defined there $\endgroup$ – Atul Mishra Jun 28 '17 at 10:53
  • $\begingroup$ I am pretty sure that it didn't include $0$. Either the sum was over Naturals or it was like $\sum_{\mathbb{Z}-{0}}$ $\endgroup$ – user456218 Jun 28 '17 at 10:55
  • $\begingroup$ Wolfram Alpha says that the sum converges to $- \frac{\pi^2}{12}$ (wolframalpha.com/input/…) $\endgroup$ – Toby Mak Jun 28 '17 at 11:11
  • $\begingroup$ Ignoring the obvious $n=0$ case, this is a well known sum (associated with Euler), so the "trick" is to remember the value, and possibly one of the derivations (the sum is without alternation is more common, but you can collect even terms, take out 1/4 and combine the result into what you want). $\endgroup$ – orion Jun 28 '17 at 11:32
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You correctly pointed out that ignoring the $n = 0$ term is not a good way to solve the problem. Either you must define what the symbol $\frac{1}{0}$ represents or you can let the series be summed over $n \in \mathbb{Z}$ such that $n \neq 0$.

In the latter case, we have $\sum_{n \in \mathbb{Z} \setminus \{0\} } \frac{(-1)^n}{n^2}$. Since the value of $n^2$ is the same independently of whether $n$ is negative or positive and any negative integer can be written in the form $-n$ for a suitable $n \in \mathbb{N}$, we can rewrite the sum as \begin{align} \sum_{n \in \mathbb{Z} \setminus \{0\} } \frac{(-1)^n}{n^2} & = \sum_{n \in \mathbb{N}} \frac{(-1)^{-n}}{n^2} + \sum_{n \in \mathbb{N}} \frac{(-1)^n}{n^2} \end{align} We know that $(-1)^{-n} = (\frac{1}{-1})^n = (-1)^n $, thus \begin{align} \sum_{n \in \mathbb{Z} \setminus \{0\} } \frac{(-1)^n}{n^2} & = \sum_{n \in \mathbb{N}} \frac{(-1)^n}{n^2} + \sum_{n \in \mathbb{N}} \frac{(-1)^n}{n^2} \\ & = 2 \sum_{n \in \mathbb{N}} \frac{(-1)^n}{n^2} \end{align} which you have already written (again, assuming $ n = 0$ is not included in your sum). This series converges due to the alternating series test since $|\frac{(-1)^n}{n^2}| > | \frac{(-1)^{n+1}}{(n+1)^2} | $ for all $n \in \mathbb{N}$ and $\lim_{n \rightarrow \infty} \frac{(-1)^n}{n^2} = 0$. With a little bit of work we can show that $\sum_{n \in \mathbb{N}} \frac{(-1)^n}{n^2} = -\frac{\pi^2}{12}$ (which you can verify using some software, e.g. Wolfram). We conclude that $\sum_{n \in \mathbb{Z} \setminus \{0\} } \frac{(-1)^n}{n^2} = -\frac{\pi^2}{6}$.

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I just figured it out: Looking at the operator $(3)$ in the original question and from the left hand side of $(1)$ the $f_{0,0}$ term is $\ell_\xi^2/24$. So from $(2)$ and since the operator that we are interested in has $\nu=n=0$ $$g_{0,0}=\sum_{\mu\in\mathbb{Z}}f_{\mu N,0}.$$ We thus have $$g_{0,0}=\frac{\ell_\xi^2}{24}+\sum_{\mu\in\mathbb{Z}\setminus\{0\}}f_{\mu N,0}=\frac{\ell_\xi^2}{24}+\frac{\ell_\xi^2}{4\pi^2}\sum_{\mu\in\mathbb{Z}\setminus\{0\}}\frac{(-1)^{\mu N}}{(\mu N)^2}.$$ And since the function in the summand in even we have $$g_{0,0}=\frac{\ell_\xi^2}{24}+\frac{\ell_\xi^2}{2\pi^2}\sum_{\mu=1}^\infty\frac{(-1)^{\mu N}}{(\mu N)^2}=\frac{\ell_\xi^2}{24}+\frac{\ell_\xi^2}{2\pi^2N^2}\text{Li}_2\left((-1)^N\right)$$ where $\text{Li}$ is the PolyLog function, therefore for even and odd $N$, it can be evaluated.

Thank you all for your help!

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