Calculating $\sum_{n \geq 0} n^2 x^n$

Question

My attempt so far:

$$\sum_{n \geq 0} n^2 x^n = x \sum_{n \geq 0} n^2 x^{n-1} = x \sum_{n \geq 0} n (x^n)'$$

And now I've got stuck. How can I continue from now?

Possible duplicate of How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$ — Clement C., Feb 21 '16 at 21:27
(Not exactly the same question, but the same ideas, technique, and straightforwardly adaptable to this version.) — Clement C., Feb 21 '16 at 21:28

score 2 · Answer 1 · answered Feb 25 '16 at 11:14

2

Hint

Using the fact that $n^2=n(n-1)+n$, then $$\sum n^2x^n=\sum n(n-1)x^n+\sum nx^n=x^2\sum n(n-1)x^{n-2}+x\sum nx^{n-1}$$ where you recognize the second and first derivatives of $\sum x^n=\frac{1}{1-x}$.

I am sure that you can take it from here.

answered Feb 25 '16 at 11:14

Claude Leibovici

260,315

score 0 · Accepted Answer · answered Feb 21 '16 at 20:56

0

Hints:

$$\sum_{n \geq 0} n^2 x^n = \sum_{n \geq 0} n(n + 1) x^{n-1} - \sum_{n \geq 0} (n + 1) x^n + \sum_{n \geq 0} x^n$$

answered Feb 21 '16 at 20:56

runaround

2,210

Calculating $\sum_{n \geq 0} n^2 x^n$

2 Answers2