Proof of 2nd Derivative of a Sum of a Geometric Series

[hmph]

Homework Statement

I am trying to prove how \(g''(r)=\sum\limits_{k=2}^\infty ak(k-1)r^{k-2}=0+0+2a+6ar+\cdots=\dfrac{2a}{(1-r)^3}=2a(1-r)^{-3}\).

I don't know what I am doing wrong and am at my wits end.

The Attempt at a Solution

(The index of the summation is always k=2 to infinity)

Ʃak(k-1)r^(k-1)
=a Ʃ k(k-1)r^(k-2)
=a Ʃ (r^k)''
=a (r^2 / (1-r)''

From this point I get a mess, and the incorrect answer. The thing I have a problem is I think Ʃ (r^k)'' when k is from 2 to infinity is r^2/(1-r), since the first term in this sequence is r^2. I don't think that is correct though

 

[SammyS]

Homework Statement

I am trying to prove how g''(r)=\sum\limits_{k=2}^\infty ak(k-1)r^{k-2}=0+0+2a+6ar+\cdots=\dfrac{2a}{(1-r)^3}=2a(1-r)^{-3}.

I don't know what I am doing wrong and am at my wits end.

The Attempt at a Solution

(The index of the summation is always k=2 to infinity)

Ʃak(k-1)r^(k-1)
=a Ʃ k(k-1)r^(k-2)
=a Ʃ (r^k)''
=a (r^2 / (1-r)''

From this point I get a mess, and the incorrect answer. The thing I have a problem is I think Ʃ (r^k)'' when k is from 2 to infinity is r^2/(1-r), since the first term in this sequence is r^2. I don't think that is correct though

To fix your LaTeX expression, I inserted the correct tags, itex and /itex in [ ] brackets.

Since the sum starts at k=2, you would not have those first two zero terms. It would be:

g''(r)=\sum\limits_{k=2}^\infty ak(k-1)r^{k-2}=2a+6ar+\cdots=\dfrac{2a}{(1-r)^3}=2a(1-r)^{-3}

So, what is it that you're supposed to be proving ?

 

[hmph]

it turns out that what i was doing was actually correct