Closed Form of Power Series

• Yagoda
In summary, the conversation discusses how to find the closed form of the series \sum_{n=0}^{\infty} n^2x^n by using the theorem for finding derivatives of power series and potentially applying the geometric series identity. The conversation concludes with suggestions to manipulate the series for g'(x) and g''(x) to find the desired closed form.

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.

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.

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?

Last edited:
Yagoda said:
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?