Thread: Series

$\displaystyle \displaystyle\sum_{r=1}^{n}(\frac{1}{r} - \frac{1}{r+1})$

How would I go about working this out? Any help appreciated. Thanks in advance.

Originally Posted by Air

$\displaystyle \displaystyle\sum_{r=1}^{n}(\frac{1}{r} - \frac{1}{r+1})$

How would I go about working this out? Any help appreciated. Thanks in advance.

$\displaystyle \displaystyle\sum_{r=1}^{n}(\frac{1}{r} - \frac{1}{r+1})$

This is a telescoping sum and so equal to:

$\displaystyle S(n)=1-\frac{1}{n+1}$

(Google for telescoping sum or series to get a detailed explanation.)

RonL

Originally Posted by CaptainBlack

$\displaystyle \displaystyle\sum_{r=1}^{n}(\frac{1}{r} - \frac{1}{r+1})$

This is a telescoping sum and so equal to:

$\displaystyle S(n)=1-\frac{1}{n+1}$

(Google for telescoping sum or series to get a detailed explanation.)

RonL

Is that the same as method of differences? If not, how can method of differences be used?

Originally Posted by Air

Is that the same as method of differences? If not, how can method of differences be used?

According to Wikipedia the answer is yes.

RonL