0
$\begingroup$

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!!

$\endgroup$
  • $\begingroup$ 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 '16 at 20:40
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Von Neumann Sep 20 '16 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.