Closed form of a sum $ \sum_{i=1}^{n}\frac{1}{((i-1)x)^2+y^2}$

Question

Consider a sum:

$$ \sum_{i=1}^{n}\frac{1}{((i-1)x)^2+y^2}$$

with $x$ and $y$ being (non-zero) constants. Is it possible to obtain a nice closed form of this expression?

Answer 1

If we consider the method explained here to prove $$ \sum_{n\geq 1}\frac{1}{n^2+1} = \frac{\pi\coth \pi-1}{2} $$ we may easily tweak it to prove also that $$ \sum_{n\geq 0}\frac{1}{(nx)^2+y^2} = \frac{1}{y^2}\sum_{n\geq 0}\frac{1}{1+n^2\left(\frac{x}{y}\right)^2} = \color{red}{\frac{1}{2y^2}+\frac{\pi \coth\left(\frac{\pi y}{x}\right)}{2 x y}}.$$ So, even if the partial sums do not have a nice closed form, the whole series does.

Answer 2

------ Complementing Jack's answer -----

Actually, since we have the formula for the infinite summation $$ S\left( {x,y} \right) = \sum\limits_{0\, \leqslant \,k} {\;\frac{1} {{\left( {x + k} \right)^{\,2} + y^{\,2} }}} = \frac{1} {{2y^{\,2} }} + \frac{{\pi \coth \left( {\pi y/x} \right)}} {{2xy}} $$ then, for a partial one we will have $$ \begin{gathered} S\left( {x,y,n} \right) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\;\frac{1} {{\left( {x + k} \right)^{\,2} + y^{\,2} }}} = \hfill \\ = \sum\limits_{0\, \leqslant \,k} {\;\frac{1} {{\left( {x + k} \right)^{\,2} + y^{\,2} }}} - \sum\limits_{n\, \leqslant \,k} {\;\frac{1} {{\left( {x + k} \right)^{\,2} + y^{\,2} }}} = \hfill \\ = \sum\limits_{0\, \leqslant \,k} {\;\frac{1} {{\left( {x + k} \right)^{\,2} + y^{\,2} }}} - \sum\limits_{0\, \leqslant \,k} {\;\frac{1} {{\left( {\left( {x + n} \right) + k} \right)^{\,2} + y^{\,2} }}} = \hfill \\ = S\left( {x,y} \right) - S\left( {x + n,y} \right) \hfill \\ \end{gathered} $$

------ in another way -----

$$ \begin{gathered} S\left( {x,y,n} \right) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\;\frac{1} {{\left( {x + k} \right)^{\,2} + y^{\,2} }}} = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\;\frac{1} {{\left( {x + k} \right)^{\,2} - \left( {iy} \right)^{\,2} }}} = \hfill \\ = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\;\frac{1} {{\left( {x + iy + k} \right)\left( {x - iy + k} \right)}}} = \frac{1} {{2iy}}\left( {\sum\limits_{0\, \leqslant \,k\, \leqslant \,n - 1} {\;\frac{1} {{\left( {x - iy + k} \right)}}} - \sum\limits_{0\, \leqslant \,k} {\;\frac{1} {{\left( {x + iy + k} \right)}}} } \right) = \hfill \\ = \frac{1} {{2iy}}\left( {\psi ^{\left( 0 \right)} \left( {x - iy + n} \right) - \psi ^{\left( 0 \right)} \left( {x + iy + n} \right) + \psi ^{\left( 0 \right)} \left( {x + iy} \right) - \psi ^{\left( 0 \right)} \left( {x - iy} \right)} \right) = \hfill \\ = \frac{1} {y}\left( {\operatorname{Im} \left( {\psi ^{\left( 0 \right)} \left( {x + iy} \right)} \right) - \operatorname{Im} \left( {\psi ^{\left( 0 \right)} \left( {x + iy + n} \right)} \right)} \right) \hfill \\ \end{gathered} $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 1}^{n}{1 \over \bracks{\pars{k - 1}x}^{2} + y^{2}} & = {1 \over xy}\,\Im\sum_{k = 0}^{n - 1}{1 \over k - \pars{y/x}\ic} = {1 \over xy}\,\Im\sum_{k = 0}^{\infty} \bracks{{1 \over k - \pars{y/x}\ic} - {1 \over k + n - \pars{y/x}\ic}} \\[5mm] & = \bbx{\ds{-\,{\Im\pars{H_{n - 1 - y\ic/x} - H_{-1 - y\ic/x}} \over xy}}} \end{align} where $\ds{H_{z}}$ is an Harmonic Number.

Answer 4

As already wrote, there is no nice closed form. A way to see an approximation is the following. Using Abel's summation we have $$S_{N}\left(z\right)=\sum_{n\leq N}\frac{1}{1+n^{2}z^{2}}=\frac{N}{1+N^{2}z^{2}}+2z^{2}\int_{1}^{N}\frac{\left\lfloor t\right\rfloor t}{\left(1+t^{2}z^{2}\right)^{2}}dt $$ where $\left\lfloor t\right\rfloor $ is the floor function and since $t-1<\left\lfloor t\right\rfloor \leq t $ we get $$S_{N}\left(z\right)<\frac{N}{1+N^{2}z^{2}}+2z^{2}\left(\int_{1}^{N}\frac{1}{1+t^{2}z^{2}}-\frac{1}{\left(1+z^{2}t^{2}\right)^{2}}dt\right) $$ $$=\frac{\arctan\left(Nz\right)}{z}-\frac{\arctan\left(z\right)}{z}+\frac{1}{z^{2}+1} $$ and in the same way we get $$S_{N}\left(z\right)>\frac{N}{1+N^{2}z^{2}}+2z^{2}\int_{1}^{N}\frac{t^{2}-t}{\left(1+t^{2}z^{2}\right)^{2}}dt $$ $$=\frac{\arctan\left(Nz\right)}{z}-\frac{\arctan\left(z\right)}{z}+\frac{1}{1+N^{2}z^{2}} $$ and $$\sum_{n=0}^{N-1}\frac{1}{y^{2}+n^{2}x^{2}}=\frac{1}{y^{2}}+\frac{1}{y^{2}}\sum_{n=1}^{N-1}\frac{1}{1+n^{2}\left(\frac{x}{y}\right)^{2}} $$ $$=\frac{1}{y^{2}}+\frac{S_{N-1}\left(\frac{x}{y}\right)}{y^{2}}.$$ For the complete series another interesting method is the following $$\sum_{n\in\mathbb{Z}}f\left(n\right)=-\sum\left\{ \textrm{residues of }\pi\cot\left(\pi z\right)f\left(z\right)\textrm{ at }f\left(z\right)\textrm{'s poles}\right\} $$ and observing that $$\frac{1}{y^{2}}\sum_{n=1}^{\infty}\frac{1}{1+n^{2}\left(\frac{x}{y}\right)^{2}}=\frac{1}{2y^{2}}+\frac{1}{2y^{2}}\sum_{n\in\mathbb{Z}}\frac{1}{1+n^{2}\left(\frac{x}{y}\right)^{2}} $$ and since we have poles at $\pm\frac{iy}{x} $ we get $$\sum_{n\in\mathbb{Z}}\frac{1}{1+n^{2}\left(\frac{x}{y}\right)^{2}}=\frac{y\pi\coth\left(\pi y/x\right)}{x} $$ and so $$\frac{1}{y^{2}}\sum_{n=1}^{\infty}\frac{1}{1+n^{2}\left(\frac{x}{y}\right)^{2}}=\color{red}{\frac{1}{2y^{2}}+\frac{\pi\coth\left(\pi y/x\right)}{2yx}}.$$