Associated Topics || Dr. Math Home || Search Dr. Math

### Telescoping Series

```
Date: 12/30/96 at 21:28:01
From: connie
Subject: Variations in progressions

I am having problems understanding the following example, which
involves the subject of varation in progression, and where the area is
denoted by "??":

Find the sum (to the nth term):

Sn =  1/(1x3) + 1/(3x5) + 1/(5x7) +....+ 1/{(2n-1)(2n+1)}

1        1
Ak =  1/{(2k-1)(2k+1)} = 1/2 ( ------ - ------ )
2k-1     2k+1

When k = 1,     1     1    1     1
--- = --- (--- - ---)
1x3    2    1     3

When k = 2,     1     1    1     1
--- = --- (--- - ---)
3x5    2    3     5

When k = n,          1         1      1         1
------------ = --- ( ------ - ------)
(2n-1)(2n+1)    2     2n-1     2n+1

?? Adding both sides of the equation, we get

1        1         n
??      Sn = --- (1- ------) = ------
2       2n+1      2n+1
```

```
Date: 12/31/96 at 05:54:21
From: Doctor Pete
Subject: Re: Variations in progressions

Hi,

I agree with you that the explanation is confusing....it took me a few
minutes to see what it meant.  What's happening here is that you have
several equations.  In each equation, the lefthand side is one
particular term of the series Sn.  On the righthand side is an
equivalent way of writing this term.  So

k=1 (first term):         1/(1x3) = (1/1-1/3)/2
k=2                       1/(3x5) = (1/3-1/5)/2
k=3                       1/(5x7) = (1/5-1/7)/2
...                       ...
k=n (last term): 1/((2n-1)(2n+1)) = (1/(2n-1)-1/(2n+1))/2.
-------------------------------------------------------------
^^^^         ^^^^^
Sum of this
column = Sn    =    Sum of this column

Clearly, if we add all these equations up, from k=1 to k=n, the
lefthand side becomes Sn, and the righthand side is:

1  [  1     1     1     1            1      1  ]
--- [ --- - --- + --- - --- + ... + ---- - ---- ]
2  [  1     3     3     5          2n-1   2n+1 ] .

Notice that every term drops out except for 1/1 and -1/(2n+1), for
that was the purpose in rewriting the individual terms.  Therefore:

Sn = (1/1 - 1/(2n+1))/2 = n/(2n+1)

I hope that makes sense.  It's a tricky idea called a *telescoping
series*; if you've ever seen a pocket telescope, like the ones
captains of sailing vessels used to have to navigate the seas, you'll
recognize that a telescoping series has the same feature of
"collapsing" down into a shorter size.

-Doctor Pete,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search