# Closed Form of Power Series

Yagoda

## Homework Statement

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

## The Attempt at a Solution

In this series, $a_n = n^2$ and $x_0 = 0$. Applying the theorem I get that $f'(x) =\sum_{n=0}^{\infty} (n+1)(n+1)^2(x)^n = \sum_{n=0}^{\infty} (n+1)^3(x)^n$. 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.

Homework Helper
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?

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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.
$$n^2 x^n = x \frac{d}{dx}\left( x \frac{d}{dx} x^n \right).$$