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 ?

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.
