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?

Since the other answer doesn't use induction, here's an induction proof.
Base case: We use the smallest value for $n$ and check that the form works.  The smallest value is $n=1$, in this case, the sum on the LHS is
$$
\sum_{k=1}^1\frac{1}{k(k+1)}=\frac{1}{1(1+1)}=\frac{1}{2}.
$$
Also, the RHS is
$$
\frac{1}{1+1}=\frac{1}{2}.
$$
Since the LHS and the RHS are equal, the claim holds when $n=1$.
Inductive case: We assume that the claim is true for $n=m$ and prove the claim is true for $n=m+1$.  Therefore, we have assumed that
$$
\sum_{k=1}^m\frac{1}{k(k+1)}=\frac{m}{m(m+1)}
$$
and we can use this fact as if it were true.
We want to prove that the statement is true when $n=m+1$, in other words, when 
$$
\sum_{k=1}^{m+1}\frac{1}{k(k+1)}=\frac{m+1}{(m+1)(m+2)}.
$$
We can't use this statement because it's the next one (the one we want to prove).
We can, however, use the inductive hypothesis to build up to the next step.  Since 
$$
\sum_{k=1}^m\frac{1}{k(k+1)}=\frac{m}{m(m+1)}
$$
we observe that the sum on the LHS is almost the sum that we need for the next case.  What's missing from the sum on the LHS is the $m+1$-st term.  In other words, we are missing the case where we plug in $m+1$ for $k$ which is $\frac{1}{(m+1)(m+1)}$.  
WARNING: The OP added $(m+1)$ to both sides and not the $m+1$-st term.  
If we add this term to both sides we get
$$
\frac{1}{(m+1)(m+2)}+\sum_{k=1}^m\frac{1}{k(k+1)}=\frac{m}{m(m+1)}+\frac{1}{(m+1)(m+2)}.
$$
The LHS is just the sum 
$$
\sum_{k=1}^{m+1}\frac{1}{k(k+1)}$$
and the RHS simplifies to the fraction
$$
\frac{m+1}{(m+1)(m+2)}.
$$
This proves the claim when $n=m+1$.
Then, by the PMI, the claim is true for all integers $n\geq 1$.
