# Does this double-sum converge?

This is a very simple question, but I don't know if this is correct or not. Is the following sum

$$\displaystyle \sum_{\substack{\ \ k,l \in \mathbb{Z} \backslash \{0\}} \\ {\ \ \ \ \ \ \ k,l \neq 0} \\ \ \ \ \ \ \ {k+l = 0}} \frac{1}{|k||l|}$$

convergent? On the one hand, it seems that we can split this into two harmonic series, which diverge. On the other hand, since $k = -l,$ is this sum equal to:

$$\displaystyle \sum_{\substack{\ \ k \in \mathbb{Z} \backslash \{0\}} \\ {\ \ \ \ \ \ \ k \neq 0}} \frac{2}{|k|^2},$$

or is this false?

• Your result is correct, if you leave out the $2$ or replace $\mathbb{Z}$ with $\mathbb{N}$ – b00n heT Nov 14 '16 at 22:35
• "On the one hand, it seems that we can split this into two harmonic series, which diverge" No idea what you mean there. Anyway, the series converges. – Did Nov 14 '16 at 22:44

Since we have the condition $k + l = 0,$ then putting $l = -k$ tells us that the sum is just
$$\displaystyle \sum_{\substack{\ \ k \in \mathbb{Z} \backslash \{0\}} \\ {\ \ \ \ \ \ \ k \neq 0}} \frac{1}{|k|^2},$$
or equivalently $2\zeta(2).$