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 $1$.
 A: You're right that in the end $$\lim_{N\rightarrow\infty}\left(\frac{1}{N}-\frac{1}{N+1}\right)=0$$ But this is not the sum itself, for this, note that $$\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots\\=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\dots$$
A: 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.
A: See the first two brackets. $-\frac{1}{2}$ in the first bracket and $\frac{1}{2}$ in the second bracket cancel each other out. This canceling happens all through out.
A: The partial sum must be:
$$S_N=\sum_{n=1}^N \left( \frac{1}{n(n+1)} \right)=\sum_{n=1}^N \left( \frac{1}{n}-\frac{1}{n+1} \right)=\\
\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\;...\color{red}{\frac1N-\frac{1}{N+1}}= \\
\left( \color{blue}{\frac{1}{1}}-\require{cancel}\cancel{\frac{1}{2}} \right)+\left( \cancel{\frac{1}{2}}-\cancel{\frac{1}{3}} \right) + \;...+ \left( \cancel{\frac{1}{N}}-\color{blue}{\frac{1}{N+1}} \right)=\\
\color{blue}1-\color{blue}{\frac{1}{N+1}}$$
Hence:
$$\lim_{N\to\infty} S_N=\lim_{N\to\infty} \left(1-\frac1{N+1}\right)=1.$$
