0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Your result is correct, if you leave out the $2$ or replace $\mathbb{Z}$ with $\mathbb{N}$ $\endgroup$ – b00n heT Nov 14 '16 at 22:35
  • $\begingroup$ "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. $\endgroup$ – Did Nov 14 '16 at 22:44
0
$\begingroup$

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 $2\zeta(2).$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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