Finding the Closed Form of a Power Series

[Yagoda]

Homework Statement

Using that \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n for |x|<1 and that
f&#039;(x) =\sum_{n=0}^{\infty} (n+1)a_{n+1}(x-x_0)^n, write \sum_{n=0}^{\infty} n^2x^n in closed form.

Homework Equations

The Attempt at a Solution

In this series, a_n = n^2 and x_0 = 0. Applying the theorem I get that f&#039;(x) =\sum_{n=0}^{\infty} (n+1)(n+1)^2(x)^n = \sum_{n=0}^{\infty} (n+1)^3(x)^n. I know I want to try to apply the sum of the geometric series and then integrate to get f(x) (or maybe those things in reverse order), but the (n+1)^3 is giving me trouble.

 

[CompuChip]

I think you failed to define f(x) - is that 1/(1 - x) or a general function $f(x) = \sum_{n = 0}^\infty a_n x^n$ ?

Consider this:
$$\sum_{n = 0}^\infty n^2 x^n = \sum_{n = 1}^\infty n^2 x^n = \sum_{n = 0}^\infty (n + 1)^2 x^{n + 1} = x \sum_{n = 0}^\infty (n + 1) (n + 1) x^n$$

and try taking it from there.

 

[Yagoda]

I am considering f(x) to be the closed form of the series that I'm looking for so I use the theorem to find its derivative f'(x) and then hopefully find f itself. So what I've got so far is that given f'(x), f(x) =\sum_{n=0}^{\infty} (n+1)^2(x)^n, which tell us that f(x) =\sum_{n=0}^{\infty} (n+1)^2(x)^n =\sum_{n=0}^{\infty} n^2x^n since that was the original series, if that makes sense.
Maybe this is an elementary question, but is there a way to apply the geometric series identity to your expression to get a closed form or does some other type manipulation that I'm not seeing have to be done?

 

[Ray Vickson]

I am considering f(x) to be the closed form of the series that I'm looking for so I use the theorem to find its derivative f'(x) and then hopefully find f itself. So what I've got so far is that given f'(x), f(x) =\sum_{n=0}^{\infty} (n+1)^2(x)^n, which tell us that f(x) =\sum_{n=0}^{\infty} (n+1)^2(x)^n =\sum_{n=0}^{\infty} n^2x^n since that was the original series, if that makes sense.
Maybe this is an elementary question, but is there a way to apply the geometric series identity to your expression to get a closed form or does some other type manipulation that I'm not seeing have to be done?

n^2 x^n = x \frac{d}{dx}\left( x \frac{d}{dx} x^n \right).
Alternatively, if $g(x) = \sum x^n,$ look at the series for $g'(x)$ and $g''(x).$ Can you see how to get $\sum n^2 x^n$ from those two series?