I've been stuck with calculating the close form of series of the following problem. Can you help me?
$\sum_{k=1}^\infty (\tan^{-1}(k) - \tan^{-1}(k+x))$
for real constant $x\geq 1$. I know that, by the comparison test, the series converges.
$$\sum_{k\geq 1}\arctan(k+x)-\arctan(k)=\sum_{k\geq 1}\int_{k}^{k+x}\frac{dz}{1+z^2}=\int_{0}^{x}\sum_{k\geq 1}\frac{1}{1+(z+k)^2}\,dz$$ equals $$ \text{Im}\int_{0}^{x}\psi(1+i+z)\,dz=\text{Im}\left[\log\Gamma(1+i+x)-\log\Gamma(1+i)\right]=\color{red}{\text{Arg}\left(\frac{\Gamma(1+i+x)}{\Gamma(1+i)}\right)} $$ which of course simplifies if $x\in\mathbb{N}$ due to $\Gamma(h+1)=h\Gamma(h)$.