Could I get an explanation for the telescoping series?

Question

I am confused on how to solve a $S_N$ using a telescoping series. I don't know where the $\frac{1}{N+1}$ comes from.

The problem reads,

Let S = $\sum\limits_{n=1}^\infty (\frac{1}{n} - \frac{1}{n+2})$

and the result should be $S_N = \frac {3}{2} - \frac {1} {N+1} - \frac {1} {N+2}$

Answer 1

Expand the sum and you will notice that absolutely all the terms except the first two and the last one cancel each other out (this is not super very easy to see, by the way):

$$\require{cancel} \sum\limits_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+2}\right)= \left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+ \left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\\ \left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+\cdots+ \left(\frac{1}{n} - \frac{1}{N+2}\right)=\\ \left(1-\cancel{\frac{1}{3}}\right)+\left(\frac{1}{2}-\cancel{\frac{1}{4}}\right)+ \left(\cancel{\frac{1}{3}}-\cancel{\frac{1}{5}}\right)+\left(\cancel{\frac{1}{4}}-\cancel{\frac{1}{6}}\right)+\\ \left(\cancel{\frac{1}{5}}-\cancel{\frac{1}{7}}\right)+\left(\cancel{\frac{1}{6}}-\cancel{\frac{1}{8}}\right)+\cdots+ \left(\cancel{\frac{1}{n}} - \frac{1}{n+2}\right)=\\ 1+\frac{1}{2}- \frac{1}{n+2}=\frac{3}{2}- \frac{1}{n+2}\xrightarrow{n\rightarrow+\infty}\frac{3}{2}-0=\frac{3}{2}. $$

So, this series adds up to $\frac{3}{2}$.

I think the user herb steinberg has got it right. I'll just expand on his answer a little bit:

$$\require{cancel} S_N=\sum\limits_{n=1}^N\left(\frac{1}{n} - \frac{1}{n+2}\right)= \sum\limits_{n=1}^N\frac{1}{n} - \sum\limits_{n=1}^N\frac{1}{n+2}= \sum\limits_{n=1}^N\frac{1}{n} - \sum\limits_{n=3}^{N+2}\frac{1}{n}=\\ \left(1+\frac12+\cancel{\frac13}+\cancel{\frac14}+\cancel{\frac15}+\cdots+\cancel{\frac1N}\right)+\left(- \cancel{\frac13}-\cancel{\frac14}-\cancel{\frac15}+\cdots-\cancel{\frac1N}-\frac{1}{N+1}-\frac{1}{N+2}\right)=\\ 1+\frac12-\frac{1}{N+1}-\frac{1}{N+2}= \frac32-\frac{1}{N+1}-\frac{1}{N+2}. $$

This is a much better way to see how the cancellation process takes place.

Answer 2

$S_N=\sum_{n=1}^N(\frac{1}{n}-\frac{1}{n+2})$ The second term can be rewritten as $\sum_{n=3}^{N+2}\frac{1}{n}$ Combine the two parts to get $S_N=1+\frac{1}{2}-\frac{1}{N+1}-\frac{1}{N+2}\to \frac{3}{2}$, as $N \to \infty$.

Answer 3

To get a closed form of $\sum\limits_{n=1}^N (\frac{1}{n} - \frac{1}{n+2})$ we write down the some first and some last summands.

$$\frac{1}{1} - \frac{1}{1+2}+\frac{1}{2} - \frac{1}{2+2}+\frac{1}{3} - \frac{1}{3+2}+\frac{1}{4} - \frac{1}{4+2}+\frac{1}{5} - \frac{1}{5+2}+...$$

$$ +\frac{1}{N-3} - \frac{1}{N-1}+\frac{1}{N-2} - \frac{1}{N}+\frac{1}{N-1} - \frac{1}{N+1}+\frac{1}{N} - \frac{1}{N+2}$$

$$=\color{blue}{\frac{1}{1}} - \frac{1}{3}+\color{blue}{\frac{1}{2}} - \frac{1}{4}+\frac{1}{3} - \frac{1}{5}+\frac{1}{4} - \frac{1}{6}+\frac{1}{5} - \frac{1}{7}+...$$

$$ +\frac{1}{N-3} - \frac{1}{N-1}+\frac{1}{N-2} - \frac{1}{N}+\frac{1}{N-1}\color{blue}{ - \frac{1}{N+1}}+\frac{1}{N} \color{blue}{- \frac{1}{N+2}}$$

The blue terms are left over. This is $\frac32- \frac{1}{N+1}-\frac{1}{N+2}$

Answer 4

Just write $$\sum\limits_{n=1}^N (\frac{1}{n} - \frac{1}{n+2}) = \sum\limits_{n=1}^N (\frac{1}{n} -\frac{1}{n+1} + \frac{1}{n+1} -\frac{1}{n+2}) = $$ $$\underbrace{\sum\limits_{n=1}^N (\frac{1}{n} -\frac{1}{n+1})}_{=1- \frac{1}{N+1}} + \underbrace{\sum\limits_{n=1}^N(\frac{1}{n+1} -\frac{1}{n+2})}_{=\frac{1}{2} - \frac{1}{N+2}} = \frac{3}{2}- \frac{1}{N+1} - \frac{1}{N+2}$$