Induction on the equality $ \sum_{i=2}^{n+1} \frac{1}{i(i+1)}= \frac{n}{2(n+2)}$ Prove with induction principle: $$ \sum_{i=2}^{n+1} \frac{1}{i(i+1)}= \frac{n}{2(n+2)} \quad \forall n \geq 2\in \mathbb{N} $$

Starting prove $P(0)$:
 $$ \sum_{i=2}^{3} \frac{1}{i(i+1)}= \frac{1}{2(2+1)}+\frac{1}{2(3+1)}= \frac{1}{4} $$
 $$ \frac{n}{2(n+2)}= \frac{2}{2(2+2)}=\frac{1}{4} $$
$\Rightarrow P(0) \text{ is true}$

Assuming $P(n)$ true:
$$ \sum_{i=2}^{n+1} \frac{1}{i(i+1)}= \frac{n}{2(n+2)}$$
Prove that $P(n+1)$ is true:
$$ \sum_{i=2}^{n+2} \frac{1}{i(i+1)}= \frac{n+1}{2(n+3)} $$
$$ \sum_{i=2}^{n+1} \frac{1}{i(i+1)}+\frac{1}{(n+2)(n+3)}= \frac{n+1}{2(n+3)} $$
from induction hypotesis $P(n)$:
$$ \frac{n}{2(n+2)}+\frac{1}{(n+2)(n+3)}= \frac{n+1}{2(n+3)} $$
$$ \frac{n(n+3)+2}{2(n+2)(n+3)} = \frac{n+1}{2(n+3)}$$
If I made everything well, from this point I have no idea how to continue, please can anyone give me some hint about. thanks in advance

you have $$\frac{n}{2(n+2)}+\frac{1}{(n+2)(n+3)}=\frac{n^2+3n+2}{2(n+2)(n+3)}=\frac{(n+2)(n+1)}{2(n+2)(n+3)}=\frac{n+1}{2(n+3)}$$
