Showing that $ \sum_{n=1}^{\infty} \arctan \left( \frac{2}{n^2} \right) =\frac{3\pi}{4}$
I would like to show that:
$$ \sum_{n=1}^{\infty} \arctan \left( \frac{2}{n^2} \right) =\frac{3\pi}{4}$$
We have:
$$ \sum_{n=1}^N \arctan \left( \frac{2}{n^2} \right) =\sum_{n=1}^N \arctan (n+1)-\arctan(n-1) =-\frac{\pi}{4}+\arctan N+\arctan(N+1)\rightarrow \frac{3\pi}{4}$$
Do you agree with my proof?
Solution 1:
Everything is fine with this proof.