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Re: How to compute this sum?


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!
|
| 



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