Summation problem

@Ian Fowler I would still be interested in seeing your approach.

@Louis Romao

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

@Louis Romao

For adding up an series of the form:

S = 1 + 2R + 3R^2 + 4R^3 + 5R^4 + .... + [nR^(n-1)] which has n terms

You will notice that each term is a derivative of R^m ( m=1 ... n)

Polynomials are continuous and differentiable, so we can re-write S as

S = dR/dR + d(R^2)/dR + d(R^3)/dR + d(R^4)/dR + d(R^5)/dR + ...... + d(R^n)/dR

Since differentiation is linear then

S = d/dR [ R + R^2 + R^3 + R^4 + R^5 + ....... +R^n]

= d/dR [ R(R^n - 1 ]/[R-1]

= d/dR [R^(n+1) - R]/[R -1]

Using the Quotient Rule and simplifying:

S = [nR^(n+1) - (n+1)R^n + 1] / [R-1]^2

If n ----> infinity when -1<R<1, we get the long division infinite series you gave. So this is really just your solution in a different guise. But it can also be used for finite series when |R| >1