Calculate $\sum_{k=1}^{n} \frac{1}{1+k^2}$

Question

Calculate the sum: $$\sum_{k=1}^{n} \frac{1}{1+k^2}$$

I'm supposed to calculate it without using functions like Gamma, Zeta, Digamma, etc...

What I tried:

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

Answer 1

The partial sums of $\sum_{k\geq 1}\frac{1}{k^2+1}$ have no simple closed form other than $\sum_{k=1}^{n}\frac{1}{1+k^2}$. On the other hand the value of the series can be computed in a rather elementary way. We may consider that for any $k\in\mathbb{N}^+$ $$ \frac{1}{k^2+1} = \int_{0}^{+\infty}\frac{\sin(kx)}{k}e^{-x}\,dx $$ holds by integration by parts. Since $$ \sum_{k\geq 1}\frac{\sin(kx)}{k} $$ is the $2\pi$-periodic extension of the function $w(x)$ which equals $\frac{\pi-x}{2}$ on $(0,2\pi)$, we have: $$ \sum_{k\geq 1}\frac{1}{k^2+1} = \int_{0}^{+\infty}w(x)e^{-x}\,dx = \sum_{m\geq 0}\int_{2m\pi}^{2(m+1)\pi}w(x)e^{-x}\,dx =\sum_{m\geq 0}e^{-2m\pi}\int_{0}^{2\pi}\frac{\pi-x}{2}e^{-x}\,dx.$$ By computing the very last integral it follows that $$ \sum_{k\geq 1}\frac{1}{k^2+1} = \left[\frac{\pi-1}{2}+\frac{\pi+1}{2}e^{-2\pi}\right]\sum_{m\geq 0}e^{-2m\pi}= \left[\frac{\pi-1}{2}+\frac{\pi+1}{2}e^{-2\pi}\right]\frac{e^{2\pi}}{e^{2\pi}-1}$$ or $$ \sum_{k\geq 1}\frac{1}{k^2+1} = \left[\pi\cosh(\pi)-\sinh(\pi)\right]\frac{1}{e^{\pi}-e^{-\pi}}=\color{red}{\frac{\pi}{2}\coth(\pi)-\frac{1}{2}}.$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 1}^{n}{1 \over 1 + k^{2}} & = \sum_{k = 1}^{n}{1 \over \pars{k + \ic}\pars{k - \ic}} = -\,{1 \over 2\ic}\sum_{k = 1}^{n}\pars{{1 \over k + \ic} - {1 \over k - \ic}} \\[5mm] & = -\Im\sum_{k = 0}^{n - 1}{1 \over k + 1 + \ic} = -\Im\sum_{k = 0}^{\infty}\pars{{1 \over k + 1 + \ic} - {1 \over k + n + 1 + \ic}} \\[5mm] & = -\Im\Psi\pars{n + 1 + \ic} + \Im\Psi\pars{1 + \ic} \\[5mm] & = \bbox[15px,#ffc,border:1px solid navy]{-\Im\Psi\pars{n + 1 + \ic} - {1 \over 2} + {1 \over 2}\,\pi\coth\pars{\pi}} \end{align}

$\ds{\Psi}$ is the Digamma Function. See $\color{black}{\bf 6.3.13}$ and $\color{black}{\bf 6.3.16}$ in this link.