Prove $\sum_{k=1}^n \frac{1}{k(k+1)} = 1 - \frac{1}{n+1}$ by induction I'm not sure if my induction proof is correct. If anyone could kindly review it, I'd appreciate it.
Prove that $n \in \mathbb{N}$:
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm+\frac{1}{n(n+1)}=1-\frac{1}{n+1}$
For $n=1$ the statement is true because $\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$
Let's assume that the statement is true for some natural number $k$, then:
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm+\frac{1}{k(k+1)}=1-\frac{1}{k+1}$
Therefore,
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}
$$
$$=1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
$$=1+\frac{1}{k+1}\cdot(\frac{-k-1}{k+2})$$
$$=1-(\frac{1}{k+1})\cdot\frac{k+1}{k+2}$$
$$=1-\frac{1}{(k+1)+1}$$

There aren't  a mistake in your third equalty. So you proof is right, clean and Nice, good work Enzo.
For $n=1$ the statement is true because $\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$
Here you are Right
Let's assume that the statement is true for some natural number $k$, then:
$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm+\frac{1}{k(k+1)}=1-\frac{1}{k+1}$
Your inductive hypotesis looks right.
Here is the critical step
Therefore,
$$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dotsm\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}
$$
$$=1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
well at this part
$$=1+\frac{1}{k+1}\cdot(\frac{-k-1}{k+2})$$
Here there aren't any mistake (Thanks to @Nem49 for notice my fault)  so you approach is correct, clean and valid  (I don't look that you simplify the expression so well and basically you do a nice work in your proof. Good work Enzo and continue working hard in your questions as now you show you big effort and welcome to Math SE. I'm apologize with you and the community ).
The way that you chose Is right and done, but other way is proceed as follow (which is more work that the work that you already do done , but is other approach for get the same result ).
You can also proceed as follow:
You know by the hypotesis
$$=1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$
$$=\frac{(k+1)(k+2)}{(k+1)(k+2)}-\frac{(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$$
$$=\frac{k^2+3k+2-k-2+1}{(k+1)(k+2)}$$
$$=\frac{k^2+2k+1}{(k+1)(k+2)}$$
$$=\frac{(k+1)^2}{(k+1)(k+2)}$$
$$=\frac{k+1}{k+2}$$
$$=\frac{k+2-1}{k+2}$$
$$=\frac{k+2}{k+2}-\frac{1}{k+2}$$
$$=1-\frac{1}{(k+1)+1}$$
And we are done.
