<s>[INST] 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 $1$.
 [/INST] 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:
$$\lim_{N\to\infty} S_N=\lim_{N\to\infty} \left(1-\frac1{N+1}\right)=1.$$
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