Find Sum To Infinity

[THUDandBLUNDER]

(1 + d) + 2(1 + 2d)r + 3(1 + 3d)r ² + 4(1 + 4d)r ³ + ......... + n(1 + nd)r ^n-1 ....... when lrl < 1

(One I thought of myself.)

[Trojan Horse]

Clarification needed: do you want the infinite sum, or just the sum to n terms? The way you've written it, it looks like you just want the sum to n terms. But given the title of the puzzle, and the fact that you said |r| must be less than 1, I'm guessing you want the infinite sum.

(I'm too lazy to work it out either way. But still... )

[THUDandBLUNDER]

[ralphmerridew]

S = sum(n = 1 to inf: n(1+nd)*r^(n-1))
rS = sum(n = 1 to inf: n(1+nd)*r^n)
rS = sum(n = 2 to inf: (n-1)*(1+(n-1)d)*r^(n-1))
S - rS = 1*(1+d)*r^0 + sum(n = 2 to inf: (n(1+nd) - (n-1)*(1+(n-1)d)) r^(n-1))
S(1 - r) = 1+d + sum(n = 2 to inf: (n+nnd - (n-1)*(1+nd-d)) r^(n-1))
S(1 - r) = 1+d + sum(n = 2 to inf: (n+nnd - n -nnd+nd + 1+nd-d) r^(n-1))
S(1 - r) = 1+d + sum(n = 2 to inf: (1 + 2nd - d) r^(n-1))
S(1 - r)r = r(1+d) + sum(n = 2 to inf: (1 + 2nd - d) r^(n))
S(1 - r)r = r(1+d) + sum(n = 3 to inf: (1 + 2(n-1)d - d) r^(n-1))
S(1 - r) - S(1 - r)r = 1+d + sum(n = 2 to inf: (1 + 2nd - d) r^(n-1)) - (r(1+d) + sum(n = 3 to inf: (1 + 2(n-1)d - d) r^(n-1)))
S(1 - r)^2 = 1+d + (1+4d - d)*r + sum(n = 3 to inf: (1 + 2nd - d) r^(n-1)) - r(1+d) - sum(n = 3 to inf: (1 + 2(n-1)d - d) r^(n-1))
S(1 - r)^2 = 1+d - r(1+d) + (1+3d)*r + sum(n = 3 to inf: ((1 + 2nd - d) - (1 + 2(n-1)d - d)) r^(n-1))
S(1 - r)^2 = 1+d-2rd + sum(n = 3 to inf: 2d r^(n-1))
S(1 - r)^2 = 1+d-2rd + 2d / (1-r)
S = (1 + d - 2rd) / (1-r)^2 + 2d / (1-r)^3

Unless I made a mistake in the algebra, which is quite likely.

[THUDandBLUNDER]

I think your first term is a bit out.
Luckily, there is a quicker way:

I obtained the given series by differentiating the following AGP term-by-term wrt r:
1 + (1 + d)r + (1 + 2d)r ² + (1 + 3d)r ³ + ....

So the sum to infinity of the series ought to be the differential coefficient of the AGP's sum to infinity.

Hence we need to differentiate 1/(1 - r) + dr/(1 - r) ² wrt r.

This gives 1/(1 - r) ² + d(1 + r)/(1 - r) ³
which equals
[(d - 1)(r + 1) + 2]/(1 - r) ³
.