Show that the series converges and find its sum Show that 
$$ \sum_{n=1}^\infty \left( \frac{1}{n(n+1)} \right) = \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+ \;... $$ 
converges and find its sum.
My solution so far:
I am thinking about finding the partial sum first and show that the series converges since its finite partial sum converges.
Now
$$ S_N=\sum_{n=1}^N \left( \frac{1}{n(n+1)} \right)=\sum_{n=1}^N \left( \frac{1}{n}-\frac{1}{n+1} \right)=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\;...= \left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right) + \;...+ \left( \frac{1}{N}-\frac{1}{N+1} \right)$$
but I don't know how to go on with this. Now $ \lim_{N \to\infty} \left( \frac{1}{N}-\frac{1}{N+1} \right)=0$ but the right answer should be $1$.

You can use the Cauchy condensation test to show convergence:
$$\sum_{n=1}^\infty \frac{2^n}{2^n(2^n+1)} = \sum_{n=1}^\infty \frac{1}{2^n+1} \le \sum_{n=1}^\infty \frac{1}{2^n} = 1$$
because the latter is a geometric series.
Hence $\displaystyle\sum_{n=1}^\infty \frac1{n(n+1)}$ also converges.
