# Calculate in closed form $\sum_{n=1}^{\infty} \frac{\arctan(1/n) H_n}{n}$

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

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

• Any reason you expect a closed form? Seems like a pretty random sum. Jun 26, 2015 at 22:43
• @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. Jun 26, 2015 at 22:45
• 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... Jun 26, 2015 at 22:59
• 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? Jun 27, 2015 at 1:56

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

• Good progress (+1) Jul 16, 2015 at 16:12
• @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 $[0,1]$. Jul 16, 2015 at 16:27
• Yes, but it's a very nice start with some good ideas to continue. Jul 16, 2015 at 18:12

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 $$\int_0^{\infty}\frac{\sin x}{x} \left(\mathrm{Li}_2(e^{-x})+\frac12\log^2(1-e^{-x})\right )dx$$

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.