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Playing with Taylor series is not helpful enough. What else would you try out?

$$\sum_{n=1}^{\infty} \frac{\arctan(1/n) H_n}{n}$$ $$\approx 2.1496160413898356727147400526167103602143301206321$$ It's easy to see the series converges since $\arctan(1/n) \approx 1/n$ when $n$ large. Maybe its integral representation makes us feel more comfortable

$$1/4\int_0^1 \frac{ 2(\gamma \pi x \coth (\pi x)+\gamma) +i x \left(\psi ^{(0)}(-i x)^2-\psi ^{(0)}(i x)^2-\psi ^{(1)}(-i x)+\psi ^{(1)}(i x)\right)}{ x^2} \, dx$$

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  • $\begingroup$ Any reason you expect a closed form? Seems like a pretty random sum. $\endgroup$
    – Winther
    Jun 26, 2015 at 22:43
  • $\begingroup$ @Winther I calculated similar (the similar word is discussable) series in closed form. I was curious to receive some clever ideas, strategies, not downvotes, or reasons for that I should not expect a closed form if some consider this is the case here. $\endgroup$ Jun 26, 2015 at 22:45
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    $\begingroup$ One should never expect a closed form if the summand is non-trivial imo, but then again some really crazy sums do have closed forms. Trying your numerical value on OEIS gives no matches. If anyone can solve your integral it's Cleo... $\endgroup$
    – Winther
    Jun 26, 2015 at 22:59
  • $\begingroup$ Do either $\displaystyle\sum_{n=1}^{\infty}\frac{\arctan(1/n)}{n}$ or $\displaystyle\sum_{n=1}^{\infty}\frac{\ln(1+1/n)H_n}{n}$ possess a closed form? $\endgroup$
    – Lucian
    Jun 27, 2015 at 1:56

3 Answers 3

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We just need to compute: $$ \mathcal{J}(m)=\sum_{n\geq 1}\frac{H_n}{n^{2m}} \tag{1} $$ but Euler's theorem (see Flajolet and Salvy, $(2.2)$) gives: $$ \mathcal{J}(m)= (1+m)\,\zeta(2m+1)-\frac{1}{2}\sum_{k=1}^{2m-2}\,\zeta(k+1)\zeta(2m-k)\tag{2}$$ as a consequence of: $$ \text{Res}\left[\left(\psi(-s)+\gamma\right)^2,s=n\right]=2 H_n,\tag{3}$$ hence: $$ S=\sum_{n\geq 1}\frac{H_n}{n}\,\arctan\frac{1}{n}=\sum_{m\geq 1}\frac{(-1)^{m+1}}{2m-1}\sum_{n\geq 1}\frac{H_n}{n^{2m}}=\\=\sum_{m\geq 1}\frac{(-1)^{m+1}}{2m-1}\left((1+m)\,\zeta(2m+1)-\frac{1}{2}\sum_{k=1}^{2m-2}\,\zeta(k+1)\zeta(2m-k)\right)=\\=\sum_{m\geq 1}\frac{(-1)^{m+1}}{2m-1}\cdot[x^{2m}]\,g(x)\tag{4}$$

where: $$\begin{eqnarray*} g(x) &=& -\gamma-\frac{\gamma x}{2}- \psi(1-x)-\gamma x \psi(1-x)-\frac{x}{2} \psi(1-x)^2+\frac{x}{2} \psi'(1-x)\\&=&\frac{\pi^2}{4}\,x+2\zeta(3)\,x^2+\frac{5}{4}\zeta(4)\,x^3+(3\zeta(5)-\zeta(2)\zeta(3))\,x^4+\ldots\tag{5}\end{eqnarray*} $$

So we have: $$ S = -\int_{0}^{1}\frac{g(ix)+g(-ix)}{2x^2}\,dx.\tag{6} $$

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  • $\begingroup$ Good progress (+1) $\endgroup$ Jul 16, 2015 at 16:12
  • $\begingroup$ @Chris'ssistheartist: there is still a lot of work to do. The RHS of $(4)$ is obviously left unchanged by replacing $g(x)$ by $\frac{g(x)+g(-x)}{2}$, that probably has a nice closed form by the reflection formulas. Then, the RHS of $(4)$ can be seen as an integral over $[0,1]$. $\endgroup$ Jul 16, 2015 at 16:27
  • $\begingroup$ Yes, but it's a very nice start with some good ideas to continue. $\endgroup$ Jul 16, 2015 at 18:12
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Start with $$\int_0^{\infty}e^{-nx}\frac{\sin x}{x}dx=\arctan \frac1n$$, (which by the way is a famous exercise in "differentiating under the integral sign" for computing $\int_0^{\infty}\frac{\sin x }{x}dx$ ). Now, $$\sum_{n=1}^{\infty}\frac{H_n}{n}x^n=-\int_0^x\frac{\log(1-t)}{t(1-t)}dt= \mathrm{Li}_2(x)+\frac12\log^2(1-x)$$ so another integral representation of the sum would be $$\int_0^{\infty}\frac{\sin x}{x} \left(\mathrm{Li}_2(e^{-x})+\frac12\log^2(1-e^{-x})\right )dx$$

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Let $f(z)=\dfrac{(\gamma+\psi_0(-z))^2}{z^2+x^2}$. On $z=Re^{i[0,2\pi]}$, $f(z)\sim\mathcal{O}\left(\dfrac{\ln^2{R}}{R^2}\right)$, so the residue theorem gives \begin{align} \sum^\infty_{n=1}\operatorname{Res}\left(f(z),n\right)+\sum_{\pm}\operatorname{Res}\left(f(z),\pm ix\right)+\operatorname{Res}\left(f(z),0\right)=0\tag1 \end{align} At the positive integers, \begin{align} \sum^\infty_{n=1}\operatorname{Res}\left(f(z),n\right) &=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\frac{1}{(z^2+x^2)(z-n)^2}+\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\frac{2H_n}{(z^2+x^2)(z-n)}\\ &=2\sum^\infty_{n=1}\frac{H_n}{n^2+x^2}-2\sum^\infty_{n=1}\frac{n}{(n^2+x^2)^2}\tag2 \end{align} At $z=\pm ix$, \begin{align} \sum_{\pm}\operatorname{Res}\left(f(z),\pm ix\right) &=\frac{(\gamma+\psi_0(-ix))^2}{2ix}-\frac{(\gamma+\psi_0(ix))^2}{2ix}\\ &=\operatorname{Im}\frac{(\gamma+\psi_0(-ix))^2}{x}\tag3 \end{align} At $z=0$, \begin{align} \operatorname{Res}(f(z),0) &=\operatorname*{Res}_{z=0}\frac{1}{z^2(z^2+x^2)}=0\tag4 \end{align} Substituting $(2)$, $(3)$, $(4)$ into $(1)$, \begin{align} \sum^\infty_{n=1}\frac{H_n}{n^2+x^2}=\sum^\infty_{n=1}\frac{n}{(n^2+x^2)^2}+\operatorname{Im}\frac{(\gamma+\psi_0(ix))^2}{2x}\tag5 \end{align} and integrating $(5)$ from $0$ to $1$ gives \begin{align} \sum^\infty_{n=1}\frac{H_n\operatorname{arccot}{n}}{n} &=\color{darkblue}{\operatorname{Im}\int^1_0\frac{(\gamma+\psi_0(ix))^2}{2x}\ {\rm d}x}+\frac{1}{2}\sum^\infty_{n=1}\frac{1}{n(n^2+1)}+\frac{1}{2}\sum^\infty_{n=1}\frac{\operatorname{arccot}{n}}{n^2}\\ &=\color{darkblue}{\mathcal{I}}+\frac{1}{2}\sum^\infty_{n=1}\left(\frac{1}{n}-\frac{1}{2(n-i)}-\frac{1}{2(n+i)}\right)+\frac{1}{2}\int^1_0\sum^\infty_{n=1}\frac{1}{n(n^2+x^2)}\ {\rm d}x\\ &=\color{darkblue}{\mathcal{I}}+\frac{\gamma}{2}+\frac{1}{2}\operatorname{Re}\psi_0(1+i)+\int^1_0\frac{\gamma+\operatorname{Re}\psi_0(1+ix)}{2x^2}{\rm d}x\\ &=\frac{\gamma}{2}+\frac{1}{2}\operatorname{Re}\psi_0(1+i)+\operatorname{Re}\int^1_0\frac{\gamma+\psi_0(1+ix)-ix(\gamma+\psi_0(ix))^2}{2x^2}\ {\rm d}x\\ &\tag6 \end{align} The remaining integral is quite hard to approach as of now.

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