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

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.
