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?
 A: This step is wrong:
$$\sum_{k=1}^{n+1} 1/k(k+1) = n/(n+1) + n+1$$
it should read:
$$\sum_{k=1}^{n+1} \dfrac{1}{k(k+1)} = \dfrac{n}{n+1} + \dfrac{1}{(n+1)(n+2)}$$
so we have:
$$\sum_{k=1}^{n+1} \dfrac{1}{k(k+1)} = \dfrac{n(n+2)+1}{(n+1)(n+2)}$$
$$= \dfrac{n^2+2n+1}{(n+1)(n+2)}$$
$$= \dfrac{(n+1)^2}{(n+1)(n+2)}$$
$$= \dfrac{n+1}{n+2}$$
A: For n=1:
$$
\sum_{k=1}^{1} 1/(k(k+1))=1/2=1/(1+1)
$$
Let's prove it for $n+1$:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}= \sum_{k=1}^{n}+\frac{1}{(n+1)(n+2)}\\
$$
But using our assumption for $P(n)$ :
$$
\sum_{k=1}^{n} \frac{1}{k(k+1)}=\frac{n}{n+1}
$$
So the sum will be:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}=\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}\\
=\frac{n(n+2)+1}{(n+1)(n+2)}\\
=\frac{n^2+2n+1}{(n+1)(n+2)}\\
=\frac{{(n+1)}^{2}}{(n+1)(n+2)}\\
=\frac{n+1}{n+2}
$$
Hence proved.
Note that you made a mistake in your last part where you wrote:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}= \frac{n}{n+1}+n+1
$$
A: $$\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1},$$
hence
$$\begin{eqnarray*} \sum_{k=1}^{n}\frac{1}{k(k+1)} &=& \sum_{k=1}^{n}\frac{1}{k}-\sum_{k=1}^{n}\frac{1}{k+1}\\&=&\sum_{k=1}^{n}\frac{1}{k}-\sum_{k=2}^{n+1}\frac{1}{k}\\&=&1-\frac{1}{n+1} = \frac{n}{n+1}.\end{eqnarray*}$$
For short: the LHS is a telescopic sum.
A: 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 $n\geq 1$.
A: 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...
