41
$\begingroup$

Is it possible to calculate the following infinite sum in a closed form? If yes, please point me to the right direction. $$\sum_{n=1}^\infty(n\ \text{arccot}\ n-1)$$

$\endgroup$
30
$\begingroup$

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

$\endgroup$
17
$\begingroup$

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

$\endgroup$
  • $\begingroup$ This is the power series of $\mathrm{arccot}(x)$ at $x=\infty$. $\endgroup$ – Mhenni Benghorbal May 16 '13 at 10:17
  • $\begingroup$ 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. $\endgroup$ – robjohn Nov 20 '13 at 8:26
8
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.