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MathGroup Archive 2005

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Re: How to compute this sum?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg63061] Re: How to compute this sum?
  • From: dh <dh at metrohm.ch>
  • Date: Tue, 13 Dec 2005 03:40:42 -0500 (EST)
  • References: <dniqkm$9r0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,
consider the denominator of your expression and convince yourself that 
it has 2 roots for n=1 and n=2
Therefore, the first 2 terms of your sum need special treatment:
First term:
Limit[(-6*2^n*n + 3*2^n + 2*3^n*n)/((n - 1)(n - 2)), n -> 1]   -->  6 - 
6*Log[3/2]
second term:
Limit[(-6*2^n*n + 3*2^n + 2*3^n*n)/((n - 1)(n - 2)), n -> 2]  --> 6*(-1 
+ Log[729/64])

Rest:
Sum[(-6*2^n*n + 3*2^n + 2*3^n*n)/(3^n)/(n^2 - 3*n + 2)/(2^n), {n, 
3,Infinity}]   --->  1/6

Therefore, your sum equals:
6 - 6*Log[3/2] + 6*(-1 + Log[729/64]) + 1/6

Daniel

mizhael wrote:
> sum((-6*2^n*n+3*2^n+2*3^n*n)/(3^n)/(n^2-3*n+2)/(2^n),n = 1 .. Inf)
> 
> thanks a lot! 
> 
> 


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