Question

Să se calculeze suma.

Answer 1

$\displaystyle S_n = \sum\limits_{k=1}^{n}\arctan\dfrac{1}{2k^2}\\ \\ \arctan x - \arctan y = \arctan\dfrac{x-y}{1+xy} \\ \\ \arctan \dfrac{1}{2k^2} = \arctan \dfrac{2}{4k^2}=\arctan\dfrac{(2k+1)-(2k-1)}{1+(2k+1)(2k-1)}= \\ \\ = \arctan(2k+1)-\arctan(2k-1) \\ \\ S_n = \sum\limits_{k=1}^n\Big[\arctan(2k+1)-\arctan(2k-1) \Big] = \\ \\ = \arctan 3+\arctan 5+...+\arctan (2n-1)+\arctan(2n+1) - \\ -\arctan 1-\arctan 3-\arctan 5-...-\arctan(2n-1) =$

$=\arctan(2n+1)-\arctan 1 = \boxed{\arctan(2n+1)-\dfrac{\pi}{4}}$