What is the sum of:

Question

$\Sigma_{n=0}^{\infty}\Sigma_{m=0}^{\infty} \frac{1}{(m-n)}$

How to treat this double sum?

For any value of $n$, the inner sum doesn't converge, even with the requirement that $m\neq n$. You essentially get the harmonic series. — saulspatz, Nov 21 '20 at 00:26
What happens when $m = n = 0$? How about when $m = n = 1$? You have infinitely many instances of this in the nested limits here... — Eric Towers, Nov 21 '20 at 00:40

score 1 · Answer 1 · answered Nov 21 '20 at 00:24

For all $n$, $$\sum_{m=0,\\ m\ne n}^\infty\frac1{m-n}=\sum_{m=0}^{n-1}\frac1{m-n}+\sum_{j=1}^\infty\frac1j=\infty$$

Therefore the double sum $\sum_{n=0}^\infty\sum_{m=0,\\ m\ne n}^\infty\frac1{m-n}$ is, in the elementary sense, $\infty$.

1 Answers1