sequences and series - Convergence of $sum_{n=1}^{infty} sum_{m=1}^{infty} frac{sin(n-m)}{n^2+m^2}$

Is the following series convergent or divergent?

$$\displaystyle{ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\sin(n-m)}{n^2+m^2}}$$

Even if it converge I do not know to prove it. However, for example, I know that
$$\sum_n \frac{\sin n}{n}$$
converges. Then, why should not converge also this double series?

Thanks in advance.

Answer

For fixed n, let $S(m,n) = \sum_{k=1}^{m}\sin (n-k).$ Then there exists $M$ such that $|S(m,n)|\le M$ for all $m,n.$ Summing by parts for fixed $n$ we find

$$\sum_{m=1}^{\infty}\frac{\sin(n-m)}{m^2+n^2} = \sum_{m=1}^{\infty}S_m[\frac{1}{m^2+n^2}- \frac{1}{(m+1)^2+n^2}]= \sum_{m=1}^{\infty}S_m\frac{2m+1}{(m^2+n^2)((m+1)^2 + n^2)}.$$

Slap absolute values everywhere in this last sum to see its absolute value is less than

$$M\sum_{m=1}^{\infty}\frac{3m}{(m^2+n^2)^2}
=\frac{1}{n^2}\sum_{m=1}^{\infty}\frac{m/n}{((m/n)^2 + 1)^2}\frac{1}{n}.$$

As $n\to \infty,$ the last sum on $m$ tends to

$$\int_0^\infty \frac{x}{(x^2+1)^2}dx < \infty.$$ It follows that there is a positive constant $C$ such that

$$\left |\sum_{m=1}^{\infty}\frac{\sin(n-m)}{m^2+n^2} \right | \le \frac{C}{n^2}$$

for all $n.$ That implies the original double sum is convergent.