TITLE: Finding $\sum_{n=1}^{\infty }\frac{243}{16(n\pi )^5}\sin(2n\pi /3)$
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\sum_{n\ =\ 1}^{\infty}{243 \over 16\pars{n\pi}^{5}}\,\sin\pars{2n\pi \over 3}}
={243 \over 16\pi^{5}}\,\Im\sum_{n\ =\ 1}^{\infty}
{\pars{\expo{2\pi\ic/3}}^{n} \over n^{5}}
={243 \over 16\pi^{5}}\,\color{#c00000}{\Im\Li{5}\pars{\expo{2\pi\ic/3}}}
\end{align}
where $\ds{\Li{s}\pars{z}}$ is the PolyLogarithm Function .

With Jonquiere Inversion Formula
  $\ds{\Li{n}\pars{\expo{2\pi\ic x}} + \pars{-1}^{n}\Li{n}\pars{\expo{-2\pi\ic x}}
=-\,{\pars{2\pi\ic}^{n} \over n!}\,\,{\rm B}_{n}\pars{x}}$ where
  $\ds{\,{\rm B}_{n}\pars{x}}$ is a
  Bernoulli Polynomial we'll have:
  \begin{align}
\color{#c00000}{\Im\Li{5}\pars{\expo{2\pi\ic/3}}}
&={\Li{5}\pars{\expo{2\pi\ic/3}} - \Li{5}\pars{\expo{-2\pi\ic/3}} \over 2\ic}
=-\pi\,{\pars{2\pi\ic}^{4} \over 5!}\,\,{\rm B}_{5}\pars{1 \over 3}
\\[5mm]&=\color{#c00000}{-\,{2\pi^{5} \over 15}\,\,{\rm B}_{5}\pars{1 \over 3}}
\end{align}
  Note that
  $\ds{\,{\rm B}_{5}\pars{x}=
     x^{5} - {5 \over 2}\,x^{4} + {5 \over 3}\,x^{3} - {1 \over 6}\,x}$ such that
  $\ds{\color{#c00000}{\,{\rm B}_{5}\pars{1 \over 3} = -\,{5 \over 243}}}$.

\begin{align}&\color{#66f}{\large%
\sum_{n\ =\ 1}^{\infty}{243 \over 16\pars{n\pi}^{5}}\,\sin\pars{2n\pi \over 3}}
={243 \over 16\pi^{5}}\bracks{\pars{-\,{2\pi^{5} \over 15}}\pars{-\,{5 \over 243}}}
=\color{#66f}{\large{1 \over 24}}
\end{align}