Prove $\frac{1}{{4n^2 - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$ Could you explain the operation in the third step?
$$\frac{1}{{4n^2  - 1}} = \frac{1}{{(2n + 1)(2n - 1)}} = \frac{1}{{2(2n - 1)}} - \frac{1}{{2(2n + 1)}}$$
It comes from the sumation $$\sum_{n=1}^\infty\frac1{4n^2-1}$$
I could just copy it to my homework, but I'd like to know how this conversion works. 
Thanks in advance.

The motivation is to write $\dfrac1{(2n+1)(2n-1)}$ as $\dfrac{A}{(2n-1)} + \dfrac{B}{(2n+1)}$. Hence, we need $A$ and $B$ such that
$$\dfrac1{(2n+1)(2n-1)} = \dfrac{A}{(2n-1)} + \dfrac{B}{(2n+1)}$$
Simplifying the right hand side, we get that
$$\dfrac1{(2n+1)(2n-1)} = \dfrac{A(2n+1) + B(2n-1)}{(2n+1)(2n-1)} = \dfrac{2(A+B)n + (A-B)}{(2n+1)(2n-1)}$$
Hence, we need $2(A+B)n + (A-B) = 1$ for all $n$. Hence, matching the coefficient of $n$ and the constant term we get that $A-B = 1$ and $2(A+B) = 0$.
This gives us $B = -\dfrac12$ and $A = \dfrac12$. Hence,
$$\dfrac1{(2n+1)(2n-1)}=\dfrac12 \dfrac1{(2n-1)} - \dfrac12 \dfrac1{(2n+1)}$$
