Presumably the sum is
\arctan\left(\frac{1}{2}\right) + \arctan\left(\frac{1}{8}\right) + \cdots + \arctan\left(\frac{1}{2n^2}\right).
If so, then what exactly does it mean to "find" this sum? If the goal is to simplify it, then this problem is similar to an old, well-known one that asks for a simplification of the sum
\sum_{k=1}^{n} \arctan\left(\frac{1}{1+k+k^2}\right).
One of the ways of doing this is to first note that \arctan(k+1) - \arctan(k) = \arctan(1/(1+k+k^2))*, and then telescope.
If you can figure out how to get this identity, then you can play around to come up with a similar one that will solve your problem.
It's also interesting to try to evaluate
\sum_{k=1}^{\infty} \arctan\left(\frac{1}{2k^2}\right).
(* What's up with ?)