calculus - 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.

$$\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$$

Answer

$$\begin{align} n\cot^{-1}(n)-1 &=n\tan^{-1}\left(\frac1n\right)-1\\ &=n\int_0^{1/n}\frac{\mathrm{d}x}{1+x^2}-1\\ &=-n\int_0^{1/n}\frac{x^2\,\mathrm{d}x}{1+x^2}\\ &=-\int_0^1\frac{x^2\,\mathrm{d}x}{n^2+x^2}\tag{1} \end{align}$$
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
$$\begin{align} \sum_{n=1}^\infty(n\cot^{-1}(n)-1) &=\frac12\int_0^1(1-\pi x\coth(\pi x))\,\mathrm{d}x\\ &=\frac12\int_0^1\left(1-\pi x\left(1+\frac{2e^{-2\pi x}}{1-e^{-2\pi x}}\right)\right)\,\mathrm{d}x\\ &=\frac{2-\pi}{4}-\pi\int_0^1\frac{xe^{-2\pi x}}{1-e^{-2\pi x}}\,\mathrm{d}x\\ &=\frac{2-\pi}{4}-\pi\int_0^1x\left(\sum_{n=1}^\infty e^{-2\pi nx}\right)\,\mathrm{d}x\\ &=\frac{2-\pi}{4}-\pi\sum_{n=1}^\infty\left(\color{#C00000}{\frac1{(2\pi n)^2}}-\left(\color{#00A000}{\frac1{2\pi n}}+\color{#0000FF}{\frac1{(2\pi n)^2}}\right)e^{-2\pi n}\right)\\ &=\frac{2-\pi}{4}-\color{#C00000}{\frac\pi{24}}-\color{#00A000}{\frac12\log\left(1-e^{-2\pi}\right)}+\color{#0000FF}{\frac1{4\pi}\mathrm{Li}_2\left(e^{-2\pi}\right)}\\ &=\frac12+\frac{17\pi}{24}-\frac12\log\left(e^{2\pi}-1\right)+\frac1{4\pi}\mathrm{Li}_2\left(e^{-2\pi}\right)\tag{3} \end{align}$$