Differences MethodDate: 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/ |
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