Sum of series with addition: $\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$

Hint:

$$\dfrac1{n^2(n+1)}=\dfrac{n+1-n}{n^2(n+1)}=\dfrac1{n^2}-\dfrac1{n(n+1)}$$

$$\dfrac1{n(n+1)}=\dfrac{n+1-n}{n(n+1)}=?$$

See Telescoping series

HINT

$$\frac{1}{n^2(n+1)}=\frac1{n^2}+\frac1{n+1}-\frac1{n}$$

Very simple $\dfrac{1}{n^2(n+1)} =\dfrac{(n+1)-n}{n^2(n+1)} =\dfrac{1}{n^2} -\dfrac{1}{n(n+1)} $

Can you take it from here?