Show that the series converges and find its sum

Question

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 $1$.

Answer 1

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 $$\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots\\=1+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+\dots$$

Answer 2

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.

Answer 3

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.

Answer 4

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: $$\lim_{N\to\infty} S_N=\lim_{N\to\infty} \left(1-\frac1{N+1}\right)=1.$$