Can we find a closed form for the series:

$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{\arctan n}{n^2}$$

Here is some basic manipulation I did:

\begin{align*} \sum_{n=1}^{\infty} \frac{\arctan n}{n^2} &=\sum_{n=1}^{\infty} \frac{1}{n^2} \sum_{k=0}^{\infty} (-1)^{k-1} \frac{n^{2k+1}}{2k+1} \\ &= \sum_{k=0}^{\infty}\frac{(-1)^{k-1}}{2k+1} \sum_{n=1}^{\infty} \frac{n^{2k+1}}{n^2}\\ &= \sum_{k=0}^{\infty} \frac{(-1)^{k-1} \zeta(1-2k)}{2k+1} \end{align*}

Now unfortunately I don't know how to proceed further. Maybe the sum does not admit a closed form and all that I have done so far be in vain. Another way would be :

$$\sum_{n=1}^{\infty} \frac{\arctan n}{n^2} =\mathfrak{Im} \left ( \sum_{n=1}^{\infty} \frac{\ln (1+in)}{n^2} \right )$$

Now I don't know how to sum the second series! Any help?

  • 2
    $\begingroup$ The series has very slow convergence. An interesting case... $\endgroup$ – Yuriy S Jul 29 '16 at 12:42
  • $\begingroup$ How do you realize the rate of convergence if it slow or not? $\endgroup$ – Tolaso Jul 29 '16 at 12:55
  • $\begingroup$ I checked with Wolfram Alpha. When I get home, I'll check with Mathematica as well $\endgroup$ – Yuriy S Jul 29 '16 at 12:59
  • 1
    $\begingroup$ Using $\tan^{-1} n = \pi/2 - \tan^{-1}(1/n)$ and the fact that $\sum 1/n^2 = \pi^2/6 $, we see that $$S = \pi^3/12 - \sum \tan^{-1}(1/n)/n^2 = \pi^3/12 - \int_0^1 \Re( H_{i x})/x^2 dx $$ where $ H_x = \psi(x+1)+ \gamma$ is the (generalized) harmonic number. $\endgroup$ – nospoon Jul 29 '16 at 13:09
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    $\begingroup$ @Tolaso I doubt that. This series is closely related with this series. Another integral representation would be $$S= \frac{\pi^3}{12} - \int_0^{\infty} \frac{\sin x}{x} \text{Li}_2( e^{-x}) dx.$$ I really don't know whether it possesses a simple closed form. $\endgroup$ – nospoon Jul 29 '16 at 13:18

The given series admits a closed form in terms of the poly-Stieltjes constants defined here.

Proposition. One has $$ \sum_{n=1}^{\infty} \frac{\arctan n}{n^2}=\frac{\pi^3}{12}+\frac1{2i}\gamma'_1(i,0)-\frac1{2i}\gamma'_1(-i,0) $$

where $\gamma'_1(a,0)$ denotes $\displaystyle \left.\partial_b\gamma_1(a,b)\right|_{b=0}$ and where $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\frac{\log^2 \!N}2\right). $$ Proof. We are allowed to perform a term by term differentiation$^1$ and to use Theorem 2 obtaining $$ \begin{align} \sum_{n=1}^{\infty} \frac{\arctan n}{n^2}&=\sum_{n=1}^{\infty} \frac1{n^2}\left(\frac{\pi}2-\arctan \frac1n \right) \\&=\frac{\pi}2\sum_{n=1}^{\infty}\frac1{n^2}-\frac1{2i}\sum_{n=1}^{\infty}\frac1{n^2}\left(\log(n+i)-\log(n-i)\right) \\&=\frac{\pi^3}{12}+\frac1{2i}\left.\partial_b\sum_{n=1}^{\infty}\frac1{n+b}\left(\log(n+i)-\log(n-i)\right)\right|_{b=0} \\&=\frac{\pi^3}{12}+\frac1{2i}\left.\partial_b\frac{}{}\left(\gamma_1(i,b)-\gamma_1(-i,b)\right)\right|_{b=0} \\&=\frac{\pi^3}{12}+\frac1{2i}\gamma'_1(i,0)-\frac1{2i}\gamma'_1(-i,0). \end{align} $$

$^1$Assume $b\ge0$ and $n\ge1$. Each function $b \mapsto f_n(b):=\frac1{n+b}\left(\log(n+i)-\log(n-i)\right)$ is differentiable over $(-1,\infty)$, the series $\sum f_n(b)$ is uniformly (normally) convergent over $[0,\infty)$, there exists a constant $c_1$ such that: $$\left|f_n(b)\right|=\left|\frac1{n+b}\left(\log(n+i)-\log(n-i)\right)\right|\le\frac{c_1}{n^2}, \quad n \to \infty,$$ and the series $\sum f'_n(b)$ is uniformly (normally) convergent over $[0,\infty)$, there exists a constant $c_2$ such that: $$\left|f'_n(b)\right|=\left|\frac1{(n+b)^2}\left(\log(n+i)-\log(n-i)\right)\right|\le\frac{c_2}{n^3}, \quad n \to \infty.$$

Remark. The proposition above may easily be generalized, giving new closed forms for a vast family of integrals. For example, one deduces a closed form for @nospoon's integral $$ \int_0^{\infty}\frac{\sin x}{x}\text{Li}_2(e^{-x})dx=\text{Im}\:\gamma'_1(-i,0) $$ using the standard result $\displaystyle \int_0^{\infty} \frac{\sin x}{x}e^{-nx}dx=\arctan \frac1n, \, n>0.$

  • $\begingroup$ I was hoping you answered ... ! Thanks! $\endgroup$ – Tolaso Jul 29 '16 at 21:02
  • $\begingroup$ @Tolaso You are very welcome. $\endgroup$ – Olivier Oloa Jul 29 '16 at 21:07
  • $\begingroup$ A rose by any other name... Are those poly-Stieltjes constants any easier to compute than the series itself $\endgroup$ – Yuriy S Aug 1 '16 at 8:07
  • $\begingroup$ @Marco Cantarini Thank you! $\endgroup$ – Olivier Oloa Aug 1 '16 at 8:10
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    $\begingroup$ @You'reInMyEye I've no problem with your comment. I will answer by this: what is $\sum_{n=1}^\infty \left( \frac1n-\ln\left(1+\frac1n \right)\right)$? the answer is $\sum_{n=1}^\infty \left( \frac1n-\ln\left(1+\frac1n \right)\right)=\gamma$ the Euler constant... Do you accept that $\gamma=\lim_{n \to \infty}\left( \sum_{k=1}^n\frac1k-\ln n\right) $ is a closed form of $\sum_{n=1}^\infty \left( \frac1n-\ln\left(1+\frac1n \right)\right)$? Concerning your last question, yes $\frac{1}{2i}(e^{i x}-e^{-i x})$ is a closed form of $\sin x$ which turns out to be very useful in numerous situations. $\endgroup$ – Olivier Oloa Aug 1 '16 at 8:51

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