convergence of double series I am currently stuck with evaluation (numerically) of the following double series:
\begin{equation}
\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}
\sin(a_1 m)\sin(a_2 m)\sin(a_3 m)\sin(b_1 n)\sin(b_2 n)\left(\frac{1 - \exp(-(m^2 + n^2))}{m(m^2+n^2)}\right)
\end{equation}
but as computations shows this series converges very slow (if it converges at all!). So my question is: does this double series converges? And if it does, how fast?
Actually I do suspect that it diverges but I can't prove it...
Thanks in advance!
 A: When $m^2 + n^2$ is large, the contribution from $1 - \frac{\mathrm{exp}(-(m^2+n^2))}{m(m^2+n^2)}$ is roughly $1$ because the fraction goes to zero as $m^2 + n^2 \to \infty$. If your series would converge, since clearly
$$
\sum_{m,n \ge 1} \frac{\mathrm{exp}(-(m^2+n^2))}{m(m^2+n^2)}
$$
converges, it would imply that 
$$
\sum_{m \ge 1} \sum_{n \ge 1} \sin(a_1m)\sin(a_2m)\sin(a_3 m) \sin(b_1n) \sin(b_2n) = \left( \sum_{m \ge 1} \prod_{i=1}^3 \sin(a_im) \right)\left( \sum_{b \ge 1} \prod_{i=1}^2 \sin(b_im) \right)
$$
would converge. I think it becomes quite believable now that this series doesn't converge in general. I leave it up to you to find out why those two remaining factors don't converge.
EDIT : With this new completely different summation... bound it this way : 
$$
\sum_{m,n \ge 1} \left| \dots \right| \le \sum_{m,n \ge 1} \frac 1{m(m^2 + n^2)}.
$$
By doing a two-dimensional version of the integral test, show that the integral
$$
\iint_{x,y \ge 1} \frac 1{x(x^2 + y^2)} dx dy
$$
converges (hint ; polar change of variables). 
Hope that helps,
