Is there a way to sum $\sum_{n=1}^{\infty}\left[ \frac{y}{n} - \tan^{-1}\left( \frac{y}{n} \right)\right]$ I'm interested in computing the sum for $y>0$:
$$
\sum_{n=1}^{\infty}\left[ \tan^{-1}\left( \frac{y}{n} \right) - \frac{y}{n} \right]
$$
In formula (42.1.6) of Hansen's A Table of Series and Products, there is a formula:
$$
\sum_{n=-\infty\\(n\neq0)}^{\infty}\left[ \tan^{-1}\left( \frac{y}{n+x} \right) - \frac{y}{n} \right] \ = \ \tan^{-1}\big( \tanh(\pi y) \cot(\pi x) \big) - \tan^{-1}\left( \frac{y}{x} \right)
$$
In taking $x \to 0^{+}$, the above formula becomes the following:
$$
\sum_{n=-\infty\\(n\neq0)}^{\infty}\left[ \tan^{-1}\left( \frac{y}{n} \right) - \frac{y}{n} \right] \ = \ 0
$$
Which simplifies to $\sum_{n=1}^{\infty}\left[ -\tan^{-1}\left( \frac{y}{n} \right) + \frac{y}{n} \right] + \sum_{n=1}^{\infty}\left[ \tan^{-1}\left( \frac{y}{n} \right) - \frac{y}{n} \right] = 0$, which is obvious. 
Is there a way to sum this series?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\ds{\sum_{n = 1}^{\infty}\bracks{\arctan\pars{y \over n} - {y \over n}}}}  =
-\int_{0}^{y}\sum_{n = 1}^{\infty}{x^{2} \over n\pars{n^{2} + x^{2}}}\,\dd x
\\[5mm] = &\
-\,\Im\int_{0}^{y}\sum_{n = 1}^{\infty}{x \over n\pars{n - \ic x}}\,\dd x =
-\,\Im\int_{0}^{y}\sum_{n = 0}^{\infty}
{x \over \pars{n + 1}\pars{n + 1 - \ic x}}\,\dd x
\\[5mm] = &\
-\,\Im\int_{0}^{y}
x\,{\Psi\pars{1} - \Psi\pars{1 - \ic x} \over \ic x}\,\dd x\qquad
\pars{~\Psi:\ Digamma\ Function~}
\\[5mm] = &\
\Re\int_{0}^{y}
\bracks{-\gamma - \Psi\pars{1 - \ic x}}\,\dd x\qquad
\pars{~\gamma:\ Euler\!-\!Mascheroni\ Constant~}
\\[5mm] = &\
-\gamma\, y - \Re\bracks{\ic\,\ln\pars{\Gamma\pars{1 - \ic y}}} =
\bbx{-\gamma\, y + \Im\ln\pars{\Gamma\pars{1 - \ic y}}}
\end{align}
A: By the inverse Laplace transform, assuming $y\in(0,1)$,
$$ \sum_{n\geq 1}\left(\arctan\frac{y}{n}-\frac{y}{n}\right)=\int_{0}^{+\infty}\left(-y+\frac{\sin(sy)}{s}\right)\frac{ds}{e^s-1}$$
and the RHS can be represented as
$$\sum_{n\geq 1}\frac{(-1)^n y^{2n+1} \zeta(2n+1)}{(2n+1)}\tag{"odd" series}$$
which, on the other hand, does not simplify nicely like the similar
$$ \sum_{n\geq 1}\frac{(-1)^n y^{2n}\zeta(2n)}{2n} = \log\sqrt{\frac{\pi y}{\sinh(\pi y)}}\tag{"even" series}.$$
Indeed the "even" series is related to $\text{Re} \log\Gamma(1+iy)$ which simplifies thanks to the reflection formula, while your series is given by the "odd" series, related to $\text{Im}\log\Gamma(1+iy)$. For values of $y$ close to zero a numerical approximation by series acceleration is pretty simple; for very large values of $y$ Stirling's approximation is recommended. 
