How can I evaluate $\sum_{n=2}^\infty n(n-1)x^n$?

Question

How can I evaluate $\displaystyle\sum_{n=2}^\infty n(n-1)x^n$?

I know the answer thanks to Wolfram Alpha, but I'm more concerned with how I can derive that answer.

Answer 1

Hint: If $f(x)=\sum x^n$ then $f''(x)=\sum n(n-1)x^{n-2}$

You will need to attend to things like precise limits.

Answer 2

Let $$f(x)=\sum_{n=0}^\infty x^n,$$ which converges to $1/(1-x)$ when $|x|<1$. Consider the function $x^2f''(x)$: $$x^2\left(\frac{d^2}{dx^2}f(x)\right)=x^2\sum_{n=0}^\infty n(n-1)x^{n-2}=\sum_{n=0}^\infty n(n-1)x^n.$$ On the other hand, you can compute a precise closed-form expression for $x^2f''(x)$ starting from $f(x)=1/(1-x)$. Once you have dealt with rigorously showing the convergence of the series, this will give you the closed form for the sum you want (after removing the first two terms from the sum, of course).

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More generally, for a polynomial $p$, we can actually say that $$\sum_{n=0}^\infty p(n)x^n=p(xD)\sum_{n=0}^\infty x^n=p(xD)\left(\frac1{1-x}\right),$$ provided that both sides converge. Here, $D$ is the differentiation operator. If you're interested, you can read more about this kind of manipulations in the book Generatingfunctionology by Herbert Wilf.

Answer 3

We can reason$$\sum_{n\ge2}n(n-1)x^n=x^2\frac{d^2}{dx^2}\sum_{n\ge2}x^n=x^2\frac{d^2}{dx^2}\frac{x^2}{1-x}=x^2\frac{d^2}{dx^2}\left(\frac{1}{1-x}-1-x\right)=\frac{2x^2}{(1-x)^3}$$provided $|x|<1$.