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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

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