If we expand, we can use a telescoping sum to find what its infinite sum is.
\(\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\)
So, we have:
\(\displaystyle \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+.................\)
Note how they all cancel out except the 1 at the beginning. The sum is 1.
Now, in general, we have \(\displaystyle \sum_{k=1}^{n}\frac{1}{k(k+1)}=1-\frac{1}{n+1}=\frac{n}{n+1}\)
Now, use induction to show this.
