Telescoping SeriesDate: 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/ |
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