Base: n=1: $\arctan(\frac{1}{2})=\arctan(\frac{1}{2})$ Suppose for n - true. Need to prove for n+1 $\arctan(\frac{1}{2})+...+\arctan(\frac{1}{2n^2})+\arctan(\frac{1}{2(n+1)^2})=\arctan(\frac{n+1}{n+1+1})$
$\arctan(\frac{n}{n+1}+\arctan(\frac{1}{2(n+1)^2})=\arctan(\frac{n+1}{n+2})$
A neat trick to prove arctan identities like this is to use the fact that the complex number $z = x + yi$ has the argument $Arg[z] = \arctan\left(\frac{y}{x}\right)$, and that $Arg[z] + Arg[w] = Arg[zw]$ (i.e. complex multiplication involves summing the arguments). So in our case we have $$\begin{split} & \arctan\left(\frac{n}{n+1}\right) + \arctan\left(\frac{1}{2(n+1)^2}\right)\\ & = Arg[n+1 + ni] + Arg[2(n+1)^2 + i] \\ & = Arg[(n+1 + ni)(2(n+1)^2 + i)] \\ & = Arg[2(n+1)^3 - n + (2n(n+1)^2 + n+1)i] \\ & = Arg[(n+2)(2n^2 + 2n + 1) + (n+1)(2n^2 + 2n + 1)i] \\ & = \arctan\left(\frac{(n+1)(2n^2 + 2n + 1)}{(n+2)(2n^2 + 2n + 1)}\right) \\ & = \arctan\left(\frac{n+1}{n+2}\right). \end{split}$$
Let us have an alternative approach:
The general term of the series is $$T_r=\arctan\left(\dfrac{1}{2r^2}\right)= \arctan\left(\dfrac{2}{4r^2}\right) $$$$=\arctan\left(\dfrac{2}{1+(4r^2-1)}\right)= \arctan\left(\dfrac{(2r+1)-(2r-1)}{1+(2r+1)(2r-1)}\right)$$ But we know, $\arctan\left(\dfrac{A-B}{1+AB}\right)=\arctan A-\arctan B$ if $AB>-1$ which obviously holds in our case.
So $T_r=\arctan(2r+1)-\arctan(2r-1)$ and the sum telescopes as $$S_n=\sum_{i=1}^n\arctan\left(\frac{1}{2r^2}\right)=\sum_{i=1}^n \arctan(2r+1)-\arctan(2r-1) $$$$=\arctan (2n+1)-\arctan 1=\arctan\left(\frac{2n+1-1}{1+(2n+1)\cdot1}\right)$$$$=\arctan\left(\frac{2n}{2n+2}\right) =\arctan\left(\frac{n}{n+1}\right) .$$
