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 
$$1/4\int_0^1 \frac{ 2(\gamma  \pi  x \coth (\pi  x)+\gamma) +i x \left(\psi ^{(0)}(-i x)^2-\psi ^{(0)}(i x)^2-\psi ^{(1)}(-i x)+\psi ^{(1)}(i x)\right)}{ x^2} \, dx$$
 A: 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:
$$ S = -\int_{0}^{1}\frac{g(ix)+g(-ix)}{2x^2}\,dx.\tag{6} $$
A: 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 
$$\int_0^{\infty}\frac{\sin x}{x} \left(\mathrm{Li}_2(e^{-x})+\frac12\log^2(1-e^{-x})\right )dx$$
A: 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.
