Sum of series involving 1+cos(n)... $$\sum_{n=2}^\infty{1+\cos(n)\over n^2}$$ 
I justified that it converges absolutely by putting it less than to $\frac{1}{n^2}$ where $p=2>1$ meaning that it converges absolutely. Would this be a correct way to solve this problem? 
 A: The Clausen functions allow for a quick and compact way to represent the sum of the series.
$$\operatorname{Sl}_{2}{\left(x\right)}=\Re{\left[\operatorname{Li}_{2}{\left(e^{ix}\right)}\right]}=\sum_{n=1}^{\infty}\frac{\cos{\left(nx\right)}}{n^2}=\frac{\pi^2}{6}-\frac{\pi x}{2}+\frac{x^2}{4}.$$
The series in question then has the sum,
$$\begin{align}
\sum_{n=2}^{\infty}\frac{1+\cos{(n)}}{n^2}
&=-1-\cos{(1)}+\sum_{n=1}^{\infty}\frac{1+\cos{(n)}}{n^2}\\
&=-1-\cos{(1)}+\operatorname{Sl}_{2}{\left(0\right)}+\operatorname{Sl}_{2}{\left(1\right)}\\
&=-1-\cos{(1)}+\frac{\pi^2}{6}+\frac{\pi^2}{6}-\frac{\pi}{2}+\frac{1}{4}\\
&=-\cos{(1)}+\frac{\pi^2}{3}-\frac{\pi}{2}-\frac34.\\
\end{align}$$
A: $\displaystyle\sum_{n=2}^\infty\left\lvert{1+\cos n\over n^2}\right\rvert=\left\lvert\displaystyle\sum_{n=2}^\infty{2\cos^2 \dfrac{n}{2}\over n^2}\right\rvert=2\left\lvert\displaystyle\sum_{n=2}^\infty{\cos^2 \dfrac{n}{2}\over n^2}\right\rvert\leq\displaystyle\sum_{n=2}^\infty{2\over n^2}=\dfrac{\pi}{3}-2$
A: Yes, as you say, $$\sum_{k=2}^\infty\left \lvert \frac{1+\cos(n)}{n^2}\right \lvert\leq \sum_{k=2}^\infty \frac{2}{n^2}=\frac{\pi^2}{3}-2,$$
which shows absolute convergence.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\sum_{n\ =\ 2}^{\infty}{1 + \cos\pars{n} \over n^{2}}}
=-1 - \cos\pars{1} + \sum_{n\ =\ 1}^{\infty}{1 \over n^{2}}
+ \Re\sum_{n\ =\ 1}^{\infty}{\pars{\expo{\ic}}^{n} \over n^{2}}
\\[5mm]&=-1 - \cos\pars{1} + {\pi^{2} \over 6}
+ \Re\Li{2}\pars{\expo{i}}
=-1 - \cos\pars{1} + {\pi^{2} \over 6}
-\half\,{\pars{2\pi\ic}^{2} \over 2!}\,\,{\rm B}_{2}\pars{1 \over 2\pi}
\end{align}
where $\ds{\,{\rm B}_{n}\pars{x}}$ is a
Bernoulli Polynomial and $\ds{\,{\rm B}_{2}\pars{x}=x^{2} - x + {1 \over 6}}$.

\begin{align}&\color{#66f}{\large%
\sum_{n\ =\ 2}^{\infty}{1 + \cos\pars{n} \over n^{2}}}
=-1 - \cos\pars{1} + {\pi^{2} \over 6}
+\pi^{2}\,\,{\rm B}_{2}\pars{1 \over 2\pi}
\\[5mm]&=-1 - \cos\pars{1} + {\pi^{2} \over 6}
+\pi^{2}\bracks{\pars{1 \over 2\pi}^{2} - {1 \over 2\pi} + {1 \over 6}}
=\color{#66f}{\large{4\pi^{2} - 6\pi - 9 \over 12} - \cos\pars{1}}
\\[5mm]&\approx {\tt 0.4288}
\end{align}

