$\lim_{n\rightarrow \infty} ( \arctan\frac{1}{2} + \arctan \frac{1}{2.2^2} +….+ \arctan \frac {1}{2n^2})$

Question

Calculate
$$\lim_{n\rightarrow \infty} \left( \arctan\frac{1}{2} + \arctan \frac{1}{2.2^2} +….+ \arctan \frac {1}{2n^2}\right)$$

My answer: i know that $$ \sum_{n=1}^N \arctan \left( \frac{2}{n^2} \right) =\sum_{n=1}^N \arctan (n+1)-\arctan(n-1)$$

as Im not able To find the $\sum_{k=1}^{n} \arctan \frac {1}{2k^2}$

I need help,,,,,any hints /solution will be aprreciated

thanks in advance

Answer 1

$$\frac{1}{2n^2}=\frac{2}{4n^2}=\frac{2}{1+4n^2-1}=\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}$$

Therefore,

$$\arctan \left( \frac{1}{2n^2}\right) = \arctan (2n+1) - \arctan(2n-1) $$

Can you perform the telescoping sum now?