Re: How to compute this sum?
- To: mathgroup at smc.vnet.net
- Subject: [mg63059] Re: How to compute this sum?
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Tue, 13 Dec 2005 03:40:40 -0500 (EST)
- Organization: The Open University, Milton Keynes, UK
- References: <dniqkm$9r0$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
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!
>
>
Do you mean, in Mathematica syntax,
In[1]:=
Sum[(-6*2^n*n + 3*2^n + 2*3^n*n)/3^n/(n^2 - 3*n + 2)/2^n, {n, 1, Infinity}]
where the denominator of the summand is zero for n ==1 or n == 2 ?
/J.M.