Bounding a series: $\frac{\pi}{2} < \sum_{n=0}^\infty \frac{1}{n^2 + 1} < \frac{3\pi}{2} $ I have the following statement -
$$\frac{\pi}{2} <  \sum_{n=0}^\infty \dfrac{1}{n^2 + 1} < \frac{3\pi}{2} $$
So I tried to prove this statement using the integral test and successfully proved the lower bound. But when I tried to calculate the upper bound I was required to calculate the integral from -1 - $\int_{-1}^{\infty} \frac{1}{x^2 +1}\,dx$.
If someone can explain why it will be great , thanks!

Note that 
$$ \sum_{n=0}^\infty \dfrac{1}{n^2 + 1}= \sum_{n=0}^\infty\int_{n}^{n+1}\frac{1}{n^2+1}dx>\sum_{n=0}^\infty\int_{n}^{n+1}\frac{1}{x^2+1}dx=\int_{0}^\infty\frac{1}{x^2+1}dx=\frac{\pi}{2}.  $$
Also note
$$ \sum_{n=0}^\infty \dfrac{1}{n^2+1}=1+\sum_{n=1}^\infty \dfrac{1}{n^2 + 1}<1+\sum_{n=1}^\infty \dfrac{1}{n^2}=1+\frac{\pi^2}{6}<\frac{3\pi}{2}. $$
Thus
$$ \frac{\pi}{2}<\sum_{n=0}^\infty \dfrac{1}{n^2+1}<\frac{3\pi}{2}. $$
