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Using the arctangent addition formula , we see that

$\arctan\left(\dfrac{1}{2k^{2}}\right) = \arctan\left(\dfrac{2}{4k^{2}}\right) =$

$\arctan\left(\dfrac{(2k + 1) - (2k - 1)}{1 + (2k + 1)(2k - 1)}\right) = \arctan(2k + 1) - \arctan(2k - 1).$

Summing from $k = 1$ to $k = n,$ we have

$(\arctan(3) - \arctan(1)) + (\arctan(5) - \arctan(3)) + (\arctan(7) - \arctan(5)) +$

$.... + (\arctan(2n - 1) - \arctan(2n - 3)) + (\arctan(2n + 1) - \arctan(2n - 1)).$

As can be seen, this is a telescoping sum, with terms canceling pairwise, leaving us

with $\arctan(2n + 1) - \arctan(1).$ The given limit then becomes

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