summation - sum of this series: $sum_{n=1}^{infty}frac{1}{4n^2-1}$

I am trying to calculate the sum of this infinite series after having read the series chapter of my textbook: $$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}$$

my steps:

$$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\sum_{n=1}^{\infty}\frac{2}{4n^2-1}-\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=..help..=sum$$

I am lacking some important properties, I feel I am coming to the right step and cannot spit that out..

Answer

Hint: Partial Fraction decomposition:$$\frac{1}{4n^2-1}=\frac{1}{(2n-1)(2n+1)}=\frac12[\frac{1}{2n-1}-\frac{1}{2n+1}]$$

You must then compute the closed form of
$$\sum_{n=1}^k[\frac{1}{2n-1}-\frac{1}{2n+1}]$$
Can you do that?
Note that
$$\sum_{n=1}^k\frac{1}{2n-1}=\frac11+\frac13+...+\frac1{2k-1}=\frac1{2\cdot 0+1}+\frac1{2\cdot 1+1}+...+\frac1{2(k-1)+1}=\sum_{n=0}^{k-1}\frac{1}{2n+1}=\sum_{n=1}^{k}\frac1{2n+1}+\frac{1}{2\cdot 0+1}-\frac1{2k+1}$$