Proof by induction: Why am I wrong?

Question

I have this exercise where I am a little stuck.

Proof the following statement:

For all $n \in \mathbb{N}$ show that $\sum_{i=1}^n \frac{1}{i(i+1)} = 1 - \frac{1}{n+1}$

Proof.

Base case: Let $n=1$.

$\sum_{i=1}^n \frac{1}{i(i+1)} = \frac{1}{1(1+1)} = 0.5$

$1 - \frac{1}{n+1} = 1 - \frac{1}{1+1} = 0.5$

Induction step:

Show that $\sum_{i=1}^{n+1} \frac{1}{i(i+1)} = 1 - \frac{1}{n+2} $

$$\sum_{i=1}^{n+1} \frac{1}{i(i+1)} =(\sum_{i=1}^{n} \frac{1}{i(i+1)}) + \frac{1}{(n+1)(n+2)} \qquad (1) \\ = 1 - \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} \qquad (2)\\ = 1 - \frac{(n+2)-1}{(n+1)(n+2)} \qquad (3)\\ = 1 - \frac{n+1}{(n+1)(n+2)} \qquad (4)\\ = 1 - \frac{1}{n+2} \qquad (5)\\ \square.$$

So here is my question. Why is step (3) in the induction step not: $1 - \frac{(n+2)+1}{(n+1)(n+2)}$?

I know that I am wrong. But I don't know exactly why, i.e. I don't know which rules apply from step (2) to (3) in the induction step. Help is much appreciated.

Answer 1

$1-\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}=1-\frac{(n+2)}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}$. The negative sign before the second expression is attached to it so when we combine the $2nd$ and $3rd$ term, we do,$1+\frac{-(n+2)}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)}=1+\frac{-(n+2)+1}{(n+1)(n+2)}=1+\frac{-((n+2)-1)}{(n+1)(n+2)}=1-\frac{(n+2)-1}{(n+1)(n+2)}$

Answer 2

We simply have

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

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

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

Answer 3

Alternate approach $$\sum_{i=1}^n \frac{1}{i(i+1)}=\sum_{i=1}^n \frac{1}{i}-\frac{1}{i+1}$$ It's certainly true for $n=1$

Apply induction now or just use the concept of telescopic series. I hope you can continue after this.