I'm facing this series:

$$\sum_{k = 1}^{+\infty}\ e^{-t\ k^2}\sin(ak)$$

where $t\in [0, +\infty)$ and $a\in [0, 2\pi]$

Despite the possibility for that series to converge (I believe the exponential term can be a sufficient condition to make it to converge, even if I found also some convergence criterions about the general series $a_n\sin(nx)$.

What actually I would like to know, if there is a way to sum that series.

I was thinking about the Abel Plana method but I think I'm stuck on the application of it (suppose it work).

Any hint or help? Thanks!!

  • $\begingroup$ If $t >0$ then the series converges absolutely by comparison with a geometric series. Showing that it converges when $t=0$ may require Euler's formula. However in that case you will get a closed form sum. $\endgroup$
    – Vik78
    Sep 20, 2016 at 20:40

1 Answer 1


When $t > 0$, we see $$\sum^\infty_{k=1} \left \lvert e^{-t k^2} \sin(ak) \right \rvert \le \sum^\infty_{k=1} e^{-tk^2} \le \sum^\infty_{k=1} e^{-tk} = \frac{e^{-t}}{1-e^{-t}} < \infty$$ so the original sum converges absolutely (hence converges). Generally if $\sum^\infty_{n=1} a_n$ converges absolutely, then $\sum^\infty_{n=1} a_n b_n$ will converge whenever $\{b_n\}$ is bounded in absolute value (which answers your questions about similar sums).

When $t=0$, it depends on the value of $a$. For example, if $a = 0, \pi, 2\pi$ it is trivial to see that the sum converges because all terms are zero. However, if $a = \pi/2$ (for example), we have $$\sum^\infty_{k=1} \sin\left( \tfrac{k\pi}{2} \right) = 1+0+(-1)+0+1+0+(-1)+\cdots$$ which diverges by oscillation.

  • $\begingroup$ Cool! So, is there a way (even not so legal) in which I might refer the second case (the divergent one) to the Grandi's Series? Namely to sum it to $\frac{1}{2}$? Or is a different oscillation? $\endgroup$
    – Enrico M.
    Sep 20, 2016 at 21:08

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