# Compute $\sum_{i,j=1}^{\infty} \frac{(-1)^{i+j}}{i^2+j^2}$

Compute

$$\sum_{i,j=1}^{\infty} \frac{(-1)^{i+j}}{i^2+j^2}$$

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I don't think it converges absolutely, which means you have to say something about the order in which the terms are taken. – Gerry Myerson Nov 2 '12 at 11:30

## 1 Answer

Let us compute $$\lim_{s\to 1}\sum_{(m,n)\neq(0,0)}\frac{(-1)^{m+n}}{(m^2+n^2)^s}$$ since that's what you want to know anyway. It's mostly about arithmetics in $\mathbb{Z}[i]$, as $m^2+n^2=(m+in)(m-in)=:N(m+in)$. Also $(-1)^{m+n}=(-1)^{m^2+n^2}$, so we're after $$f(s)=\sum_{\alpha\in\mathbb{Z}[i]-\{0\}}\frac{(-1)^{N\alpha}}{(N\alpha)^s}.$$ Notice that $N\alpha$ is even iff $(1+i)\vert\alpha$. Using unique factorization to primes (and the fact that there are $4$ units) we get

$$\sum_{\alpha\in\mathbb{Z}[i]-\{0\}}\frac{(-1)^{N\alpha}}{(N\alpha)^s}=4(-1+2^{-s}+4^{-s}+\dots)\times\prod_\pi\frac{1}{1-(N\pi)^{-s}},$$ where $\pi$ runs over all primes in $\mathbb{Z}[i]$ except for $1+i$. Now either $N\pi=p$ where $p\equiv 1\text{ mod } 4$ is a prime, and it occurs for two $\pi$'s, or $N\pi=q^2$ where $q\equiv 3\text{ mod } 4$ is a prime (that's when $\pi=q$). As $$\frac{1}{1-2^{-s}}\prod_p\frac{1}{(1-p^{-s})^2}\prod_q\frac{1}{1-q^{-2s}}$$ $$=\zeta(s)\prod_p\frac{1}{1-p^{-s}}\prod_q\frac{1}{1+q^{-s}}$$ $$=\zeta(s)(1-3^{-s}+5^{-s}-7^{-s}+9^{-s}-\dots),$$ we have $$f(s)=4(2^{1-s}-1)\zeta(s)(1-3^{-s}+5^{-s}-7^{-s}+9^{-s}-\dots).$$ Since $\lim_{s\to1}(s-1)\zeta(s)=1$, $\lim(2^{1-s}-1)/(s-1)=-\log2$, and $$\lim_{s\to1}1-3^{-s}+5^{-s}-7^{-s}+9^{-s}-\dots=\pi/4$$ we get $$\lim_{s\to 1}f(s)=-\pi\log2.$$ (I'm amazed that I got the right answer:)

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I'm amazed, too :-) What a great answer! – joriki Nov 5 '12 at 9:06
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