Differences Method

Date: 07/06/2001 at 05:36:29
From: Mark Atkinson
Subject: The method of differences

How does the method of differences work?

Date: 07/06/2001 at 05:51:55
From: Doctor Anthony
Subject: Re: The method of differences

To find the sum of n terms and the sum to infinity of the series

   1         1         1
 ----  +  ------  + ------ + ........
 1 x 2     2 x 3     3 x 4

You use the method of differences.

   u(n) = 1/(n(n+1))      =  1/n     -  1/(n+1)
   u(n-1)= 1/((n-1)n)     =  1/(n-1) -  1/n
   u(n-2)= 1/((n-2)(n-1)) =  1/(n-2) -  1/(n-1)
    ............................................
    ............................................
   u(3) = 1/((3)(4))      =  1/3     -  1/4
   u(2) = 1/((2)(3))      =  1/2     -  1/3
   u(1) = 1/((1)(2))      =  1/1     -  1/2
  --------------------------------------------  Add all the equations
 SUM[u(r)]                = 1  - 1/(n+1)
 r = 1 to n

When you add all the equations you get cancellation between rows, and 
everything goes out except the first term of u(1) and the last term of 
u(n).

So sum of the series is  1 - 1/(n+1)

as n -> infinity the term 1/(n+1) -> 0, and so the sum of the series 
to infinity is 1.

  SUM[u(r)]        = 1
 (r = 1 to infinity)
   
- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/