I'd really appreciate some help with this one, its got me stumped. I know the answer is 4, but don't know how to deal with the r in the numerator. Appreciate any help.

I got it using a formula for what is called an Arithmetico-Geometric Series for the sum of n terms. I had to monkey around with the index to start it at 1 instead of 2, let n app. infinity and 4 pops out. But I have to say that the above method is waaay better than mine. Damn, I love it when someone comes up with a simple and elegant solution. Thanks Louis!

I'd really appreciate some help with this one, its got me stumped. I know the answer is 4, but don't know how to deal with the r in the numerator. Appreciate any help.

I got it using a formula for what is called an Arithmetico-Geometric Series for the sum of n terms. I had to monkey around with the index to start it at 1 instead of 2, let n app. infinity and 4 pops out. But I have to say that the above method is waaay better than mine. Damn, I love it when someone comes up with a simple and elegant solution. Thanks Louis!

I will leave out the R = 2 to infinity.

Factor out 1/3 to get S = 1/3*sigma [R(2/3)^(R-2)]

[1] 3S = sigma [R(2/3)^(R-2)]

= 2(2/3)^0 + 3(2/3)^1 + 4(2/3)^2 + 5(2/3)^3 + 6(2/3)^4 + .....

Now multiply both sides by 2/3

[2] (2/3)3S = 2(2/3)^1 + 3(2/3)^2 + 4(2/3)^3 + 5(2/3)^4 + .....

Now subtract: (collecting like terms - except for the first term))

3S - 2S = 2(2/3)^0 +[ (2/3) + (2/3)^2 + (2/3)^3 + (2/3)^4 + .... ]

S = 2 + (2/3)/[1 - (2/3)]

= 2 + (2/3)/(1/3)

= 2+2

= 4

Note that this approach will also work for finite series where |R| >1

Brilliant, thanks very much