Prove This Inequality ${\pi \over 2} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}} \le {\pi \over 2} + 1$ $${\pi \over 2} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}} \le {\pi \over 2} + 1$$
I see I should use Riemann sum, and that
$$\int_0^{\infty} {dx \over {1+x^2}} \le \sum_{n=0}^{\infty} {1 \over {1+n^2}}$$
But how do I explain this exactly, and how to get the other half of the inequality (with ${\pi \over 2} + 1$)?

Hint:
$$ \sum_{n=0}^{\infty} {1 \over {1+n^2}}=\frac{1}{2}+\frac{\pi}{2}\coth(\pi) $$
$$\coth(\pi)> 1\simeq 1$$
so
$$\sum_{n=0}^{\infty} {1 \over {1+n^2}}\simeq\frac{1}{2}+\frac{\pi}{2}$$
