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 ${\pi \over 2} + 1$)?
 A: Hint:
$$
\overbrace{\sum_{n=1}^\infty\frac1{1+n^2}}^{S-1}\le\overbrace{\int_0^\infty\frac{\mathrm{d}x}{1+x^2}\vphantom{\sum_{n=0}^\infty}}^{\pi/2}\le\overbrace{\sum_{n=0}^\infty\frac1{1+n^2}}^S
$$

$\displaystyle\int_0^\infty\frac{\mathrm{d}x}{1+x^2}\le\sum_{n=0}^\infty\frac1{1+n^2}$


$\displaystyle\sum_{n=1}^\infty\frac1{1+n^2}\le\int_0^\infty\frac{\mathrm{d}x}{1+x^2}$

A: Hint. For each $n=0,1,2,\cdots$, you have, for all $x \in [n,n+1]$,
$$
n^2+1\leq 1+x^2\leq (n+1)^2+1
$$ giving
$$
\frac1{(n+1)^2+1}\leq \frac1{x^2+1}\leq \frac1{n^2+1}
$$ then integrating with respect to  $x$ from $n$ to $n+1$,
$$
\int_n^{n+1}\frac1{(n+1)^2+1}\:dx\leq \int_n^{n+1}\frac1{x^2+1}\:dx\leq \int_n^{n+1}\frac1{n^2+1}\:dx,
$$ observe that
$$
\int_n^{n+1}\frac1{(n+1)^2+1}\:dx=\frac1{(n+1)^2+1},\quad \int_n^{n+1}\frac1{n^2+1}\:dx=\frac1{n^2+1}
$$ giving
$$
\frac1{(n+1)^2+1}\leq \int_n^{n+1}\frac1{x^2+1}\:dx\leq \frac1{n^2+1}
$$ then sum from $n=0$ to $+\infty$, and with a change of indice on the left, you may conclude easily using
$$
\int_0^\infty\frac1{x^2+1}\:dx=\lim_{x \to +\infty}\arctan x-\arctan 0=\frac{\pi}2.
$$
A: With the same reason:
$$\sum_{n=0}^{\infty} {1 \over {1+n^2}} \le \int_0^{\infty} {dx \over {1+(x-1)^2}}$$
$$\sum_{n=0}^{\infty} {1 \over {1+n^2}} \le \arctan(\infty)-\arctan(-1)=\frac{\pi}2+\frac{\pi}4 \le \frac{\pi}2+1$$
A: Hint You can try by using the Taylor series of arctan evaluated at x=1
A: Hint:
$$ \sum_{n=0}^{\infty} {1 \over {1+n^2}}=\frac{1}{2}+\frac{\pi}{2}\coth(\pi) $$
$$\coth(\pi)> 1\simeq 1$$
so
$$\sum_{n=0}^{\infty} {1 \over {1+n^2}}\simeq\frac{1}{2}+\frac{\pi}{2}$$
