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