# 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?

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \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}

• Hi Felix my friend. This is very similar to how I did this (and deleted after I saw your post). But I don't believe that the closed form we obtained is more useful than the original series for numerical computation. Apr 3, 2018 at 2:38
• @MarkViola Hi, Mark my friend. I saw your answer and it's right. That's true: The involved $\Gamma$ will require a numerical evaluation as the original post. Apr 3, 2018 at 2:43

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.

• I am quite interested how the last equation is obtained. Could you please provide some related references? Apr 3, 2018 at 2:27
• @Szeto: the $\log\Gamma$ function is involved and $\Gamma(1-is)\Gamma(is)$ simplifies by the reflection formula. Apr 3, 2018 at 2:32
• When $y = 1$, the result is $\approx -0.2756$ but your result yields $\approx -0.6509$. Apr 3, 2018 at 2:34
• How did you use the ILT to arrive at the first equation? Apr 3, 2018 at 2:40
• @FelixMarin: the series involving $\zeta(2n+1)$ returns the correct outcome, the second series is shown just for providing a similar example. Apr 3, 2018 at 2:41