Prove $\frac{1}{4{n}^{2}-1}=\frac{1}{(2n+1)(2n-1)}=\frac{1}{2(2n-1)}-\frac{1}{2(2n+1)}$

Could you explain the operation in the third step?

$\frac{1}{4{n}^{2}-1}=\frac{1}{(2n+1)(2n-1)}=\frac{1}{2(2n-1)}-\frac{1}{2(2n+1)}$

It comes from the sumation

$\sum _{n=1}^{\mathrm{\infty}}\frac{1}{4{n}^{2}-1}$

I could just copy it to my homework, but I'd like to know how this conversion works. Thanks in advance.

Could you explain the operation in the third step?

It comes from the sumation

I could just copy it to my homework, but I'd like to know how this conversion works. Thanks in advance.