For Nihar and Kartik

$\begin{aligned} \displaystyle \sum_{n=1}^{\infty} \arctan \left ( \dfrac{1}{2n^{2}} \right ) \\= \sum_{n=1}^{\infty} \arctan \left ( \dfrac{2}{4n^{2} +1 -1 } \right ) \\= \sum_{n=1}^{\infty} \arctan \left ( \dfrac{2}{1 + (2n^{2}-1)(2n^{2}+1) } \right ) \\= \sum_{n=1}^{\infty} \arctan (2n+1) - \arctan (2n-1) \end{aligned}$

Now this is a Telescoping series which can be simplified to ,

$\begin{aligned} \arctan (2n+1) - \arctan 1 \\ = \arctan \left ( \dfrac{2n}{1+(2n+1)(1)} \right ) \end{aligned}$

Hence , the $tan$ value of which yields ,

$\lim_{n \rightarrow \infty} \dfrac{n}{n+1} = 1$