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! > >