Using Mathematica, we claim that the following series is convergent: $$\sum_{n=1}^{\infty}\frac{\sin(n^2 t)}{n}$$ Any idea how we prove this?

  • $\begingroup$ In terms of $t$? $\endgroup$ – user285523 Nov 8 '15 at 5:26
  • $\begingroup$ It is convergent for $t=2k\pi,\,k\in\mathbb Z$ $\endgroup$ – Kamil Jarosz Nov 8 '15 at 8:32

Let $t=\frac{\pi}{2}$.

Then since the sequence of squares mod $4$ is $$\{1^2,2^2,3^2,4^2,5^2,6^2,7^2,8^2,\ldots\}\equiv\{1,0,1,0,1,0,1,0,\ldots\}$$ your series will be the same as $$\begin{align} &\frac{\sin(\pi/2)}1+\frac{\sin(0)}2+\frac{\sin(\pi/2)}3+\frac{\sin(0)}4+\frac{\sin(\pi/2)}5+\frac{\sin(0)}6+\frac{\sin(\pi/2)}7+\frac{\sin(0)}8+\cdots\\ &=\frac11+\frac13+\frac15+\frac17+\cdots \end{align}$$ which is divergent.

So at least for $t=\pi/2$, this is actually not a convergent series.

In general, for $t=2\pi/q$, if the squares mod $q$ are not balanced between "highs" and "lows", it seems likely that there will be divergence. For example, with $q=7$, the sequence of squares is $\{0,1,4,2,2,4,1,\text{repeat}\}$. So you have $$\begin{align} &\sum_{n=0}^{\infty}\left(\frac{\sin(2\pi/7)}{7n+1}+\frac{\sin(8\pi/7)}{7n+2}+\frac{\sin(4\pi/7)}{7n+3}+\frac{\sin(4\pi/7)}{7n+4}+\frac{\sin(8\pi/7)}{7n+5}+\frac{\sin(2\pi/7)}{7n+6}\right)\\ &=\sum_{n=0}^{\infty}\left(\sin(4\pi/7)\left(\overbrace{\frac{2\cos(4\pi/7)}{7n+2}+\frac{1}{7n+3}}^{\text{positive}}+\overbrace{\frac{1}{7n+4}+\frac{2\cos(4\pi/7)}{7n+5}}^{\text{positive}}\right)+\frac{\sin(2\pi/7)}{7n+1}+\frac{\sin(2\pi/7)}{7n+6}\right)\\ &>\sum_{n=0}^{\infty}\left(\frac{\sin(2\pi/7)}{7n+1}+\frac{\sin(2\pi/7)}{7n+6}\right) \end{align}$$ which is divergent.

So it seems like there are lots of $t$ for which this does not converge. Having said that, for $t$ that are not rational multiples of $\pi$, my guess (it's only a guess) would be that the sum converges, since the sign of the terms would exhibit pseudorandom behavior.

  • $\begingroup$ Yes..the series doesn't converge for many values. Will see if I can get something about when it converges $\endgroup$ – Mohammed Sababheh Nov 9 '15 at 11:46

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.