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) | | thanks a lot! | |