Is there a closed-form of $\sum_{n=1}^{\infty }\frac{\cos^2(n)}{n^2}$

Question

Is there a closed-form of $$\sum_{n=1}^{\infty }\frac{\cos^2(n)}{n^2}$$

Answer 1

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: $$\sum_{n\geq 1}\frac{\sin^2 n}{n^2}=\frac{\pi-1}{2}, \qquad \sum_{n\geq 1}\frac{\cos^2 n}{n^2}=\color{red}{\frac{\pi^2-3\pi+3}{6}}.\tag{4}$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

Answer 3

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.