Compute the series $\sum_{n=1}^{+\infty} \frac{1}{n^3\sin(n\pi\sqrt{2})}.$ I need to compute $$\sum_{n=1}^{+\infty} \frac{1}{n^3\sin(n\pi\sqrt{2})}.$$ This an exercice of "Amar and Matheron, complex analysis". I proved the convergence and now to compute the sum, I follow the hint of the book which is : Consider integrals of the form $$\int_{\gamma}\frac{dz}{z^3[\sin(\pi z)\sin(\sqrt{2}-1)\pi z]}$$ for a well-chosen $\gamma.$ I know this a residue theorem application but it seems a bit hard to have the good idea. I also tried with a summation factor. Any help will be greatly appreciated.
 A: The contour $\gamma$ you want is the square having vertices $\pm (N-1/2) (1 \pm i)$.  You can show that, as $N \to \infty$, the contour integral goes to zero.
However, the contour integral has poles at the integers and at the integers times $\sqrt{2}+1$.  The residue at $z=0$ may be evaluated by expansion in a Laurent series, as the pole here is of order $5$.  This expansion looks like
$$\frac1{z^3} \frac1{\pi z \left (1-\frac{\pi^2 z^2}{3!} + \frac{\pi^4 z^4}{5!}+\cdots \right )} \frac1{(\sqrt{2}-1)\pi z \left (1-\frac{(\sqrt{2}-1)^2\pi^2 z^2}{3!} + \frac{(\sqrt{2}-1)^4 \pi^4 z^4}{5!}+\cdots \right )}$$
The coefficient of $1/z$ in this expansion is essentially the coefficient of $z^4$ in the expansion of the sine terms in parentheses, or
$$\frac{13 \pi^2}{90} \left ( \sqrt{2}-1 \right )$$
The residue at each pole $z=n \ne 0$ is simply
$$\frac{(-1)^n}{\pi n^3 \sin{(\sqrt{2}-1) \pi n}} = \frac1{\pi n^3 \sin{\sqrt{2} \pi n}}$$
The residue at each pole $z=(\sqrt{2}+1) n \ne 0$ is
$$\frac{(-1)^n (\sqrt{2}+1)}{(\sqrt{2}+1)^3 \pi n^3 \sin{(\sqrt{2}+1) \pi n}} = \frac{(\sqrt{2}-1)^2}{\pi n^3 \sin{\sqrt{2} \pi n}}$$
Thus,
$$2 \sum_{n=1}^{\infty} \frac{1+(\sqrt{2}-1)^2}{\pi n^3 \sin{\sqrt{2} \pi n}} + \frac{13 \pi^2}{90} \left ( \sqrt{2}-1 \right ) = 0$$
because the contour integral is zero.  Thus, I get that

$$\sum_{n=1}^{\infty} \frac{1}{n^3 \sin{\sqrt{2} \pi n}} = -\frac{13 \pi^3}{360 \sqrt{2}} $$

ADDENDUM
I had some thoughts about this sum.  First of all, let's talk about its convergence, which is not trivial.  Numerical experiments are more or less helpful, but as one might expect, there is a bit of jumping around the numerical value of the result I have derived.  So, even thought the OP stated that he had proven convergence, I just want to illustrate how convergence is achieved.
At issue is the factor $\sin{\sqrt{2} \pi n}$ of each term in the sum: when is this sine term dangerously close to zero?  If we think about it for a bit, the worst-case scenario is when $2 n^2$ is one less or more than a perfect square.  (Recall that $2 n^2$ can never be a perfect square.)  That is, when
$$2 n^2 = m^2 \pm 1$$
for some $m \in \mathbb{N}$.  In this case, 
$$\sin{\sqrt{2} \pi n} = \sin{\left (\sqrt{m^2 \pm 1} \pi \right )} $$
For $n$ sufficiently large, i.e., $m$ large as well, we have
$$\sin{\sqrt{2} \pi n} \approx \sin{\left [m \pi \left (1 \pm \frac1{2 m^2} \right ) \right ]} = (-1)^m \sin{\frac{\pi}{2 m}} \approx (-1)^m \frac{\pi}{2 m}$$
Thus,
$$\left | \frac1{n^3 \sin{\sqrt{2} \pi n}} \right | \le \frac1{n^3 \frac{\pi}{2 \sqrt{2} n}} = \frac{2 \sqrt{2}}{\pi n^2}$$
and, because the worst-case scenario term is bounded by something times $1/n^2$, the series converges by comparison with the sum of $1/n^2$.  
Why is this so important?  Well, it looks like we have discovered a bug in Mathematica.  As a matter of routine, I check the result against a straight evaluation in Mathematica.  To my horror, Mathematica returned $-13 \pi^{\color{red}{2}}/(360 \sqrt{2})$.  How was I off by a factor of $\pi$?  I checked and checked my work but found nothing wrong.
The solution to this problem lay in simply evaluating the sum numerically for an increasingly large number of terms.  However, in order to assess whether there would be any surprises waiting for us from the sine term, I had to estimate the worst possible "spike" near an integer times $\pi$.  What I found above is that, worst case, the terms decrease as some constant times $1/n^2$, so the effect of any spike is limited.  
Armed with this information, I was able to verify in Mathematica that, indeed, numerical evaluations of finite sums converged to the answer I gave above rather than Mathematica's result.  Mr. Wolfram will be receiving yet another letter.
ADDENDUM II
I did send that letter, and here is what I got in response:

Hello Ron,
Thank you for taking the time to send in this report. It does appear
  that this sum is missing a factor of Pi, even in the latest version of
  Mathematica (10.3.1), and I have forwarded an incident report to our
  developers with the information you provided.
We are always interested in improving Mathematica, and I want to thank
  you once again for bringing this issue to our attention. If you run
  into any other behavior problems, or have any additional questions,
  please don't hesitate to contact Wolfram Technical Support
  (support@wolfram.com).
Sincerely,
[name redacted]

Remember, just because Mathematica or Maple says something, it is not always true.
ADDENDUM III
I just got (20 Nov 2016) the following email from the fine folks at Wolfram Research:

Hello Ron Gordon,
In Febuary 2016 you reported an issue with Mathematica wherein Sum
  returns a wrong answer for some expressions. We believe that the issue
  has been resolved in the current release of Mathematica.
Thank you for your report and we look forward to a continued,
  productive relationship with you.
Best regards,  Wolfram Technology Group Wolfram
  Research, Inc. http://www.wolfram.com/support

I have verified that this error has been corrected in Version 11.0.1.  Thanks to Wolfram Research for helping me get the latest version installed.
