Is there a closed-form of $\sum_{n=1}^{\infty }\frac{\cos^2(n)}{n^2}$ Is there a closed-form of $$\sum_{n=1}^{\infty }\frac{\cos^2(n)}{n^2}$$
 A: Since over the interval $(0,\pi)$ we have:
$$\frac{\pi-x}{2}=\sum_{n\geq 1}\frac{\sin(nx)}{n}\tag{1}$$
over the same interval we have also:
$$\frac{2\pi x-x^2}{4} = \sum_{n\geq 1}\frac{1-\cos(nx)}{n^2}\tag{2} $$
or:
$$\frac{2\pi x-x^2}{8} = \sum_{n\geq 1}\frac{\sin^2\left(n\frac{x}{2}\right)}{n^2}\tag{3} $$
so by setting $x=2$ we get:
$$\sum_{n\geq 1}\frac{\sin^2 n}{n^2}=\frac{\pi-1}{2}, \qquad \sum_{n\geq 1}\frac{\cos^2 n}{n^2}=\color{red}{\frac{\pi^2-3\pi+3}{6}}.\tag{4}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\sum_{n=1}^{\infty }\frac{\cos^{2}\pars{n}}{n^{2}}}\ =\
\overbrace{\sum_{n=1}^{\infty }\frac{1}{n^{2}}}^{\dsc{\frac{\pi^{2}}{6}}}\ -\
\sum_{n=1}^{\infty }\frac{\sin^{2}\pars{n}}{n^{2}}
=\frac{\pi^{2}}{6} + 1 - \sum_{n=0}^{\infty }\,{\rm sinc}^{2}\pars{n}
\end{align}
where $\ds{\,{\rm sinc}}$ is the Cardinal Sine Function.
With the Abel-Plana Formula:
\begin{align}&\color{#66f}{\sum_{n=1}^{\infty }\frac{\cos^{2}\pars{n}}{n^{2}}}
=\frac{\pi^{2}}{6} + 1 - \bracks{%
\overbrace{\int_{0}^{\infty }\,{\rm sinc}^{2}\pars{x}\,\dd x}^{\dsc{\frac{\pi}{2}}} \ +\  \half\,\ \overbrace{{\rm sinc}^{2}\pars{0}}^{\dsc{1}}}
\ =\ \color{#66f}{\frac{\pi^{2} - 3\pi + 3}{6}} \approx {\tt 0.5741}
\end{align}
A: You can have a closed form in terms of the polylogarithm function. Write the series as
$$ \frac{1}{4}\sum_{n=1}^{\infty} \frac{e^{2i n}}{n^2}+ \frac{1}{4}\sum_{n=1}^{\infty} \frac{e^{-2i n}}{n^2} +\frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2}  = \frac{1}{4}(Li_2(e^{2i}) + Li_2(e^{-2i})+ \zeta(2)). $$
You can simplify the above.
