How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$ How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$  ?
My Attempt :   I was using Integral Test of a Series. I got $\int_0^\infty \frac{1}{1+x^2}  \le \sum_{n=0}^\infty \frac{1}{1+n^2}  \le \frac{1}{1+0^2} + \int_0^\infty \frac{1}{1+x^2}$  which gives $ \frac{\pi}{2}  \le \sum_{n=0}^\infty \frac{1}{1+n^2}  \le 1+ \frac{\pi}{2}$.
Can anyone please help me by giving any hint ?
 A: We shall give a stronger bound. Note that
$$ 2=1+\sum^\infty_{n=1}\frac{1}{n^2+n}<\sum^\infty_{n=0}\frac{1}{n^2+1}<1+\frac{1}{2}+\sum^\infty_{n=2}\frac{1}{n^2}=\frac{1}{2}+\frac{\pi^2}{6}$$
which is a stronger bound.
A: This can be computed using the result of this answer:
$$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+1}
&=-\frac1{2i}\sum_{n=0}^\infty\left(\frac1{i-n}+\frac1{i+n}\right)\\
&=\frac12-\frac1{2i}\sum_{n\in\mathbb{Z}}\frac1{i+n}\\
&=\frac12-\frac1{2i}\pi\cot(\pi i)\\[3pt]
&=\frac12+\frac\pi2\coth(\pi)\\[9pt]
&\approx2.076674
\end{align}
$$
We can get bounds tighter than $\frac\pi2$ and $\frac{3\pi}4$ with telescoping series:
$$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+1}
&\ge1+\sum_{n=1}^\infty\frac1{n^2+n}\\
&=1+\sum_{n=1}^\infty\left(\frac1n-\frac1{n+1}\right)\\[6pt]
&=2
\end{align}
$$
and, although this does not give as tight an upper bound as in Zhang's answer, it only uses telescoping series
$$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+1}
&\le1+\frac12+\sum_{n=2}^\infty\frac1{n^2-\frac14}\\
&=1+\frac12+\sum_{n=2}^\infty\left(\frac1{n-\frac12}-\frac1{n+\frac12}\right)\\[3pt]
&=\frac{13}6
\end{align}
$$
Thus,
$$
2\le\sum_{n=0}^\infty\frac1{n^2+1}\le\frac{13}6
$$
