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MathGroup Archive 2005

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Re: How to compute this sum?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg63077] Re: How to compute this sum?
  • From: Peter Pein <petsie at dordos.net>
  • Date: Tue, 13 Dec 2005 03:41:11 -0500 (EST)
  • References: <dniqkm$9r0$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

mizhael schrieb:
> 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! 
> 
> 
First have a closer look at the roots of the denominator:

Limit[(3*2^n - 3*2^(1 + n)*n + 2*3^n*n)/(6^n*(2 - 3*n + n^2)), n -> #1]& /@ {1, 2}
--> {1 + Log[2] - Log[3], -(1/6) + Log[3/2]}

add these:
Simplify[Plus @@ %]
--> 5/6

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

add everyting:
[ left as exercise ;-), so you can do a small part of your homework by yourself ]

Peter


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