Infinite series $\sum_{n=1}^{\infty}{\log}^2(1+\frac{1}{n})$ Does anybody help me obtain a closed form for the following series (via a bit elementary ways)?
$$\sum_{n=1}^{\infty}{\log}^2\left(1+\frac{1}{n}\right)$$
Seemingly, the answer would be $$2\zeta(2)-\sum_{n=1}^{\infty} \frac {\zeta (2n)}{n^2}.$$
There are some questions with answers (e.g. this one) on Stack Exchange relating to this series and pointing out the answer, but they possess advanced approaches in this regard.
Thanks in advance!
 A: The end of the answer by Jack D'Aurizio gives the most elementary solution I see.
So let me explain it in more detail here. We use properties of $$\newcommand{\Li}{\operatorname{Li}_2}\newcommand{\substeq}[1]{\underset{#1}{\phantom{\big[}=\phantom{\big]}}}\Li(z)=-\int_0^z\frac{\log(1-t)}{t}\,dt\qquad(z\leqslant 1)\tag{1}\label{dilogdef}$$ (let's restrict the domain to the reals). Namely, the identity $$\Li(1-z)+\Li(1-z^{-1})=-\frac12\log^2 z\qquad(z>0)\tag{2}\label{diloghard}$$ is obtained from the definition \eqref{dilogdef} using substitutions: $$\Li(1-z)\substeq{t=1-x}-\int_z^1\frac{\log x}{1-x}\,dx,\quad\Li(1-z^{-1})\substeq{t=1-x^{-1}}\int_z^1\frac{\log x}{x(1-x)}\,dx$$ (added, these give $\int_z^1\frac{\log x}{x}\,dx$). Similarly, we have the identity $$\Li(z)+\Li(-z)=\frac12\Li(z^2)\qquad(|z|\leqslant 1)\tag{3}\label{dilogeasy}$$ and the power series [obtained from the one for $\log(1-t)$] $$\Li(z)=\sum_{k=1}^\infty\frac{z^k}{k^2}.\qquad(|z|\leqslant 1)\tag{4}\label{dilogseries}$$
In particular, $\Li(1)=\zeta(2)\color{gray}{[{}=\pi^2/6]}$. Now, starting with \eqref{diloghard} at $z=1+n^{-1}$, we get
\begin{align*}
\sum_{n=1}^\infty\log^2\left(1+\frac1n\right)&=-2\sum_{n=1}^\infty\left[\Li\left(-\frac1n\right)+\Li\left(\frac1{n+1}\right)\right]
\\\color{gray}{[\text{reindex terms}]}\quad&=2\Li(1)-2\sum_{n=1}^\infty\left[\Li\left(-\frac1n\right)+\Li\left(\frac1n\right)\right]
\\\color{gray}{[\eqref{dilogeasy}\text{ at }z=n^{-1}]}\quad&=2\zeta(2)-\sum_{n=1}^\infty\Li\left(\frac1{n^2}\right)
\\\color{gray}{[\eqref{dilogseries}\text{ at }z=n^{-2}]}\quad&=2\zeta(2)-\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{k^2 n^{2k}}
\\\color{gray}{[\text{sum over }n]}\quad&=2\zeta(2)-\sum_{k=1}^\infty\frac{\zeta(2k)}{k^2}.
\end{align*}
