Let's look at partial sums: $\displaystyle S_k = \sum_{n=1}^k \dfrac{1}{4n^2-1}$. We have
$\displaystyle S_1 = \dfrac{1}{3}$
$\displaystyle S_2 = \dfrac{1}{3} + \dfrac{1}{15} = \dfrac{2}{5}$
$\displaystyle S_3 = \dfrac{2}{5} + \dfrac{1}{35} = \dfrac{3}{7}$
In general, the pattern appears to be $\displaystyle S_k = \dfrac{k}{2k+1}$. You can show inductively that this is the case.
So, $\displaystyle \sum_{n=1}^\infty \dfrac{1}{4n^2-1} = \lim_{k \to \infty} \dfrac{k}{2k+1} = \dfrac{1}{2}$.