Series

**Simplicity** · Jan 26th 2008, 09:54 AM

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

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

**CaptainBlack** · Jan 26th 2008, 10:00 AM

Originally Posted by Air

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

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

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

This is a telescoping sum and so equal to:

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

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

RonL

**Simplicity** · Jan 26th 2008, 10:01 AM

Originally Posted by CaptainBlack

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

This is a telescoping sum and so equal to:

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

(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?

**CaptainBlack** · Jan 26th 2008, 10:13 AM

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

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