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