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MathGroup Archive 2005

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Re: How to compute this sum?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg63059] Re: How to compute this sum?
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Tue, 13 Dec 2005 03:40:40 -0500 (EST)
  • Organization: The Open University, Milton Keynes, UK
  • References: <dniqkm$9r0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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! 
> 
> 
Do you mean, in Mathematica syntax,

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

where the denominator of the summand is zero for n ==1 or n == 2 ?

/J.M.


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