Calculate the following infinite sum in a closed form $\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$? Is it possible to calculate the following infinite sum in a closed form? If yes, please point me to the right direction.
$$\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$$
 A: We can have the following integral representation 
$$ \sum_{n=1}^\infty(n\ \text{arccot}\ n-1)=\int _{0}^{\infty }\!{\frac {x\cos \left( x \right) -\sin \left( x
 \right) }{{x}^{2} \left( {{\rm e}^{x}}-1 \right) }}{dx} \sim - 0.4152145872, $$
which agrees with Wolfram.
A: $$
\begin{align}
n\cot^{-1}(n)-1
&=n\tan^{-1}\left(\frac1n\right)-1\\
&=n\int_0^{1/n}\frac{\mathrm{d}x}{1+x^2}-1\\
&=-n\int_0^{1/n}\frac{x^2\,\mathrm{d}x}{1+x^2}\\
&=-\int_0^1\frac{x^2\,\mathrm{d}x}{n^2+x^2}\tag{1}
\end{align}
$$
Using formula $(9)$ from this answer and substituting $z\mapsto ix$, we get
$$
\sum_{n=1}^\infty\frac{1}{n^2+x^2}=\frac{\pi\coth(\pi x)}{2x}-\frac{1}{2x^2}\tag{2}
$$
Combining $(1)$ and $(2)$ yields
$$
\begin{align}
\sum_{n=1}^\infty(n\cot^{-1}(n)-1)
&=\frac12\int_0^1(1-\pi x\coth(\pi x))\,\mathrm{d}x\\
&=\frac12\int_0^1\left(1-\pi x\left(1+\frac{2e^{-2\pi x}}{1-e^{-2\pi x}}\right)\right)\,\mathrm{d}x\\
&=\frac{2-\pi}{4}-\pi\int_0^1\frac{xe^{-2\pi x}}{1-e^{-2\pi x}}\,\mathrm{d}x\\
&=\frac{2-\pi}{4}-\pi\int_0^1x\left(\sum_{n=1}^\infty e^{-2\pi nx}\right)\,\mathrm{d}x\\
&=\frac{2-\pi}{4}-\pi\sum_{n=1}^\infty\left(\color{#C00000}{\frac1{(2\pi n)^2}}-\left(\color{#00A000}{\frac1{2\pi n}}+\color{#0000FF}{\frac1{(2\pi n)^2}}\right)e^{-2\pi n}\right)\\
&=\frac{2-\pi}{4}-\color{#C00000}{\frac\pi{24}}-\color{#00A000}{\frac12\log\left(1-e^{-2\pi}\right)}+\color{#0000FF}{\frac1{4\pi}\mathrm{Li}_2\left(e^{-2\pi}\right)}\\
&=\frac12+\frac{17\pi}{24}-\frac12\log\left(e^{2\pi}-1\right)+\frac1{4\pi}\mathrm{Li}_2\left(e^{-2\pi}\right)\tag{3}
\end{align}
$$
A: We have
$$\text{arccot}(x) = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)x^{2k+1}}\implies x\text{arccot}(x)-1 = \sum_{k=1}^{\infty} \dfrac{(-1)^k}{(2k+1)x^{2k}}$$
Hence,
$$\sum_{n=1}^{\infty}\left(n\text{arccot}(n)-1\right) = \sum_{n=1}^{\infty}\sum_{k=1}^{\infty} \dfrac{(-1)^k}{(2k+1)n^{2k}} = \sum_{k=1}^{\infty} \dfrac{(-1)^k \zeta(2k)}{2k+1}$$
