Inductive proof of series (sum of 1 / [i*(i+1)] ) I like to show that
$$
\sum_{i=1}^n \frac{1}{i(i+1)} = 1 - \frac{1}{n+1} \quad   \forall n \in \mathbb{N}.
$$
This seems to be a possible solution:
Base case:
$$ n_0 = 1 $$
$$ \sum_{i=1}^1 \frac{1}{1*2} = 1 - \frac{1}{1+1}$$
$$ \frac{1}{2} = \frac{1}{2} $$
Induction hypothesis:
$$ \sum_{i=1}^n \frac{1}{i(i+1)} = 1 - \frac{1}{n+1} $$
Induction step:
$$ \sum_{i=1}^{n+1} \frac{1}{i(i+1)} = \sum_{i=1}^{n} \frac{1}{i(i+1)} + \frac{1}{(n+1)(n+2)} $$
with the induction hypothesis
$$ = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} \quad (1) $$
$$ = 1 - \frac{(n+2) \color{red}{-1} }{(n+1)(n+2)} \quad (2) $$
$$ = 1 - \frac{n+1}{(n+1)(n+2)} \quad (3) $$
$$ = 1 - \frac{1}{n+2} \quad (4) $$
Now, my problem is that I cannot follow from equation (1) to equation (2). How it comes to "minus 1" in the numerator.
If I do this I come to
$$ 1 - \frac{(n+2) \color{red}{+1} }{(n+1)(n+2)} \quad (2a) $$
Could someone help?

$$\frac{\color{red}{-1}}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{\color{red}{-1}(n+2)+1}{(n+1)(n+2)}=\color{orange}{-}\frac{(n+2)\color{orange}{-1}}{(n+1)(n+2)}$$
