Re: How to compute this sum?
- To: mathgroup at smc.vnet.net
- Subject: [mg63073] Re: How to compute this sum?
- From: "Jens-Peer Kuska" <kuska at informatik.uni-leipzig.de>
- Date: Tue, 13 Dec 2005 03:40:58 -0500 (EST)
- Organization: Uni Leipzig
- References: <dniqkm$9r0$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
for n->1 and n->2
(n^2-3*n+2) is zero and you will have
Infinity + Infinity+
Sum[(-6*2^n*n + 3*2^n + 2*3^n*n)/(3^n)/(n^2 - 3*n
+ 2)/(2^n), {n , 3 ,
Infinity}]== Infinity+Infinity+1/6
Regards
Jens
"mizhael" <loseminds at hotmail.com> schrieb im
Newsbeitrag news:dniqkm$9r0$1 at smc.vnet.net...
|
sum((-6*2^n*n+3*2^n+2*3^n*n)/(3^n)/(n^2-3*n+2)/(2^n),n
= 1 .. Inf)
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| thanks a lot!
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