<s>[INST] $\frac{\pi}2 < \sum_0^\infty \frac{1}{1+n^2} < \frac{3\pi}4 $ Prove that:
$\frac{\pi}2 < \sum_0^\infty \frac{1}{1+n^2} < \frac{3\pi}4 $ 
What I've tried:
I solved the improper integral:  $\int_0^\infty  \frac{1}{1+x^2} = \lim_{b\to \infty} \arctan b -\arctan 0 = \frac{\pi}2 $.
Now, everything in the sum (and in the integral) is positive so the sum must be lower than the integral (it has more members), so the claim is wrong (originally it was a "Prove/Disprove" problem...). However, if we only add the first 3 members we get 1.7 which is bigger than $\frac{\pi}2 $.
Also, I checked with wolfram and the sum is somewhere around 2 so the claim is true. I have no idea how to prove it. 
Someone suggested I would read Basel's problem ,in which the integral is also smaller than the sum (starting from 1) but it led me nowhere and I don't think that this kind of complex solution is needed here. 
 [/INST] An advanced approach using zeta function values $\zeta(2)$ and $\zeta(4)$ to get even better bounds. 
For $n>1$, $$\frac{1}{1+n^2}=\frac{1}{n^2}\frac{1}{1+\frac{1}{n^2}} = \frac1{n^2}-\frac{1}{n^4}+\frac{1}{n^6}-\frac{1}{n^8}+...$$
So $$\frac{1}{n^2}-\frac{1}{n^4}<\frac{1}{n^2+1}<\frac{1}{n^2}$$
So:
$$\frac{3}{2}+\left(\zeta(2)-\zeta(4)-\left(\frac{1}{2^2}-\frac{1}{2^4}\right)\right)=1+\frac{1}{2}+\sum_{n=2}^\infty \left(\frac{1}{n^2} -\frac{1}{n^4}\right)<\sum_{n=0}^{\infty}\frac{1}{n^2+1}  < \frac{1}{2} + \zeta(2)$$
Now, $$\frac{1}{2}+\zeta(2)=\frac 12+\frac{\pi^2}{6} <\frac{3\pi}{4}$$
And:
$$\frac{21}{16}+\zeta(2)-\zeta(4) = \frac{21}{16} + \frac{\pi^2}{6} - \frac{\pi^4}{90}>\frac{\pi}{2}$$
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