What is represented by $\sum_{n=1}^{\infty} n^2 x^{n-1}$?

Question

Is it possible to represent the following series in term of a non-infinite function: $$ \sum_{n=1}^{\infty} n^2 x^{n-1} $$

If you start with If $\displaystyle f(x)= \sum_{n=1}^{\infty} x^{n} = \frac{x}{1-x}$ then $\displaystyle \frac{d}{dx} \left(x \frac{d}{dx}f(x)\right)= \sum_{n=1}^{\infty} n^2 x^{n-1} = \frac{1+x}{(1-x)^3} $ — Henry, Feb 23 '16 at 09:38

Yuriy S · Answer 1 · 2016-02-23T09:38:58.717

3

Yes

$$ \sum_{n=1}^{\infty} n^2 x^{n-1}=\frac{1}{x} \sum_{n=1}^{\infty} n^2 x^{n}=\frac{(1+x)}{(1-x)^3} $$

Converges when $|x|<1$

edited Feb 23 '16 at 09:38

answered Feb 23 '16 at 09:29

Yuriy S

I think you should divide by $x$ rather than multiply – Henry Feb 23 '16 at 09:37
@Henry, thanks, I need to get more sleep – Yuriy S Feb 23 '16 at 09:39

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