Closed form for $\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}$

Question

Find the closed form of $$\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)}$$

Answer 1

Hint: Note that $$\dfrac{2}{(2k-1)(2k+1)}=\dfrac{1}{2k-1}-\dfrac{1}{2k+1}.$$

This leads to $$\sum_{k=1}^n \dfrac{1}{(2k+1)(2k-1)}=\frac{1}{2}\sum_{k=1}^n \left( \dfrac{1}{2k-1}-\dfrac{1}{2k+1}\right).$$

Can you continue from here?