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!

  • $\begingroup$ It converges when one of the $a_i,\, b_j$ is an integer multiple of $\pi$. $\endgroup$ – Daniel Fischer Jun 28 '13 at 11:28
  • $\begingroup$ 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. $\endgroup$ – MichalisN Jun 28 '13 at 11:31
  • $\begingroup$ Sorry, there was mistake. I've just fixed it $\endgroup$ – 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).

Hope that helps,

  • $\begingroup$ Sorry... There was mistake in formulation... (( $\endgroup$ – mechanician Jun 28 '13 at 12:04
  • $\begingroup$ I answered your new question. $\endgroup$ – Patrick Da Silva Jun 28 '13 at 12:44
  • $\begingroup$ How would you formally define the Integral Test in two dimensions ? $\endgroup$ – Zophikel Feb 9 '18 at 15:31
  • 1
    $\begingroup$ @Zophikel : It's exactly the same idea as in the one-dimensional integral test. In the one-dimensional test, if you can find a function above your sequence which is integrable, decreasing and the integral is convergent, then your sum also converges (since it is smaller). Without going into too much detail, you can just think of functions which lie above your double summation in the plane where the summation is defined ; the role of the 2d rectangle replacing your sequence value becomes a 3d rectangular prism. and you want an integrable function sitting above those prisms. $\endgroup$ – Patrick Da Silva Feb 9 '18 at 15:39
  • $\begingroup$ @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. $\endgroup$ – Patrick Da Silva Feb 9 '18 at 15:41

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