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

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

 A: 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
$$ \sum_{k\geq 1}\frac{1}{k^2+1} = \left[\pi\cosh(\pi)-\sinh(\pi)\right]\frac{1}{e^{\pi}-e^{-\pi}}=\color{red}{\frac{\pi}{2}\coth(\pi)-\frac{1}{2}}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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.
