<s>[INST] Finding $\sum_{n=1}^{\infty }\frac{243}{16(n\pi )^5}\sin(2n\pi /3)$ The WolfarmAlpha couldn't give me the sum of $$\sum_{n=1}^{\infty }\frac{243}{16(n\pi )^5}\sin(2n\pi /3)$$ therefore I thought that this problem is difficult so I used my calculator to get $(1/24)$
Is this value right or not?
If this right, why the WolfarmAlph couldn't find it? 
 [/INST] Consider the sum:
$$\sum_{n=1}^{\infty} \frac{\displaystyle\sin{\frac{2 n \pi}{3}}}{n^5} $$
This sum is a bit easier than it looks, if you know the residue theorem.  The main observation is that the numerator either takes the value $\sqrt{3}/2$, $-\sqrt{3}/2$, or $0$.  Then you may rewrite the sum as
$$\frac{\sqrt{3}}{2} \sum_{n=1}^{\infty} \left [\frac1{(3 n-2)^5} - \frac1{(3 n-1)^5}\right ] = \frac{\sqrt{3}}{4 \cdot 3^5} \sum_{n=-\infty}^{\infty} \left [\frac1{(n-2/3)^5} - \frac1{(n-1/3)^5}\right ]$$
We may apply the residue theorem to this sum by considering the following result:
$$\sum_{n=-\infty}^{\infty} f(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k} [ \cot{\pi z} \, f(z) ]$$
where $z_k$ are the non-integer poles of $f$.  All we need to do is compute the residues at the poles, which in this case are at $z=1/3$ and $z=2/3$.  For example,
$$\begin{align}\operatorname*{Res}_{z=2/3} \left [ \cot{\pi z} \, \left ((z-2/3)^{-5}-(z-1/3)^{-5}  \right ) \right] &= \frac1{4!} \left [\frac{d^4}{dz^4} \cot{\pi z}   \right ]_{z=2/3} - \frac1{4!} \left [\frac{d^4}{dz^4} \cot{\pi z}   \right ]_{z=1/3} \\ &= -\frac{8\pi^4}{3 \sqrt{3}}\end{align}$$
The sum is then
$$\sum_{n=1}^{\infty} \frac{\displaystyle\sin{\frac{2 n \pi}{3}}}{n^5} = \frac{2 \pi^5}{729}$$
The result follows.
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