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...

• It converges when one of the $a_i,\, b_j$ is an integer multiple of $\pi$. – Daniel Fischer Jun 28 '13 at 11:28
• This really depends on the $a_i$. For example: If one is in $\pi\cdot\mathbb{Z}$ then the series is $0$. If they are all in $\pi\cdot\mathbb{Q}\setminus\pi\cdot \mathbb{Z}$ and $a_1=a_2$ and $b_1=b_2$ the series diverges. – MichalisN Jun 28 '13 at 11:31
• Sorry, there was mistake. I've just fixed it – mechanician Jun 28 '13 at 12:06

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).
• @Zophikel : Some conditions ensure that your choice of function $f$ indeed lies above all those prisms, for instance if the trace functions $x \mapsto f(x,y)$ are decreasing for all $y$ and the trace functions $y \mapsto f(x,y)$ are decreasing for all $x$ ; this is easily checked by looking at the sign of the partial derivatives, which is what I mentally did in my answer above. – Patrick Da Silva Feb 9 '18 at 15:41