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Compute the following sum $$\sum_{n=1}^{+\infty}\frac{1}{n^3 \sin(n \pi \sqrt{2})}$$

Source :

Very difficult sum of series

Like jmerry on AoPS I have no idea how to compute the sum.

Any ideas ? Perhaps someone knows already the result..

Thank you in advance for your time.

Jean-Claude Arbaut
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    I would say the first question is whether it converges. That sine could be very small for various $n$, making occasional big terms in the sum. – Gerry Myerson May 19 '14 at 12:47

2 Answers2

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This is not a solution, but it doesn't quite fit as a comment and with some hope it can be finished. Let $a=\pi\sqrt{2}$ and $$g_k(x)=\sum_{n\in\mathbb{Z},n\neq0}\frac{1}{n^k}\frac{\exp(inx)}{\exp(ina)-\exp(-ina)},$$ you want to know $g_3(0)$. For convergence, $1/\sin na=O(n)$, so $g_k$ converges absolutely and uniformly for $k\geq3$, and we can use $g_k'=ig_{k-1}$ to get lower $k$ (the series converges unconditionally in the sense of distributions for every $k$).

Notice now $$g_k(x+a)-g_k(x-a)=\sum_{n\neq0}\frac{\exp(inx)}{n^k}=c_kB_k((x\text{ mod }2\pi)/2\pi)\quad(*)$$ where $c_k$ is an easy constant and $B_k$ is Bernoulli polynomial. Also $g_k(-x)=(-1)^{k+1}g_k(x)$.

The difference equation $(*)$ would have a polynomial solution - if the RHS were a polynomial (not just a piecewise polynomial) and if $g_k$ didn't have to be $2\pi$-periodic. Nonetheless, I hope that it can be solved also in the periodic case, and thus finally get $g_3(0)$.

Sorry for this irresponsible suggestion.

user8268
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  • How do you get $1/\sin na=O(n^2)$? – Gerry Myerson May 20 '14 at 13:18
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    @GerryMyerson I actually made a mistake and should have written $O(n)$ (will fix it). The thing is that $\sqrt{2}$, as any quadratic irrational, can't be well approximated by a rational number, i.e. $\exists c>0$ s.t. for every $m,n\in\mathbb Z$, $n\neq0$, $|\sqrt{2}-m/n|>1/n^2$ - and thus $|n\sqrt{2}-m|>1/n$. (for proof - the continued fraction of $\sqrt{2}$ is eventually periodic). – user8268 May 20 '14 at 15:00
  • @GerryMyerson errata: "$>1/n^2$" and "$>1/n$" should have been "$>c/n^2$" and "$>c/n$" – user8268 May 20 '14 at 15:26
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Partial answer:

$$\sum_{n\ge 1}\frac{1}{n^3\sin(n\pi \sqrt{2})}=\sum_{n\ge 1}\sum_{k=0}^{\infty}\frac{e^{-jn\pi\sqrt{2}}}{2j n^3}e^{-2jkn\pi\sqrt{2}}=\frac{1}{2j}\sum_{k\ge 0}\mbox{Li}_{3}\left(e^{-j(2k+1)\pi \sqrt{2}}\right)$$ where $\mbox{Li}_s(\cdot)$ is the polylogarithm function. We need to see if there is a closed form for this kind of summation.

Samrat Mukhopadhyay
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