Proof by induction for "sum-of" Prove that for all $n \ge 1$:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$
What I have done currently:
Proved that theorem holds for the base case where n=1.
Then:
Assume that $P(n)$ is true. Now to prove that $P(n+1)$ is true:
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n}{n+1} + n+1$$
So:
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n+1}{n+2}$$
However, how do I proceed from here?

Just to address your specific problem regarding induction, there is an error in the way you have introduced the $n+1$ term. You need to add the $n+1$th term onto the sum to $n$ terms:
Assume it is true for $S_n$, so that $$S_n=\Sigma_{k=1}^{n}\frac{1}{k(k+1)}=\frac{n}{n+1}.$$
Now consider $$S_{n+1}=S_n+\frac{1}{(n+1)(n+2)}=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}$$
$$=\frac{n^2+2n+1}{(n+1)(n+2)}=\frac{\overline{n+1}}{\overline{n+1}+1}$$
then we are done...
