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

\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}

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, $$\sum_{n=1}^{\infty}\left(n\text{arccot}(n)-1\right) = \sum_{n=1}^{\infty}\sum_{k=1}^{\infty} \dfrac{(-1)^k}{(2k+1)n^{2k}} = \sum_{k=1}^{\infty} \dfrac{(-1)^k \zeta(2k)}{2k+1}$$

• This is the power series of $\mathrm{arccot}(x)$ at $x=\infty$. – Mhenni Benghorbal May 16 '13 at 10:17
• Faster convergence can be achieved with $$\frac\pi4-1+\sum_{k=1}^\infty(-1)^k\frac{\zeta(2k)-1}{2k+1}$$ which converges approximately $0.6$ digits per term. – robjohn Nov 20 '13 at 8:26

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.

• Hi. How did you get its integral representation? I'm curious about the technique used. (+1) – user 1357113 May 16 '13 at 7:15
• Im also interested – Greg.Paul Jan 3 '18 at 5:59