Find a closed form expression for $\sum\limits_{i=0}^{\infty} i^2\cdot x^i$

Question

I think I'm supposed to use calculus somewhere but I'm completely lost.

Answer 1

First, show that for $|x|<1$, $\sum\limits_{n=0}^{\infty} nx^n=\frac{x}{(1-x)^2}$. You can get this from differentiating $\frac{1}{1-x}=\sum\limits_{n=0}^{\infty} x^n$ and multiplying by $x$ (to take $x^{n-1}$ back to $x^n$ after differentiating).

Then, just repeat, i.e. differentiate $\sum\limits_{n=0}^{\infty} nx^n=\frac{x}{(1-x)^2}$ and multiply by $x$ to find $\sum\limits_{n=0}^{\infty} n^2x^n$.

Answer 2

Write $i^2=(i+2)(i+1)-3(i+1)+1$. Then:

$$\begin{align}\sum_{i=0}^\infty i^2 x^i &= \sum (i+2)(i+1)x^i -3\sum (i+1)x^i + \sum x^i\\ &=\frac{d^2}{dx^2}\left(\frac{1}{1-x}\right)-3\frac{d}{dx}\left(\frac{1}{1-x}\right) + \frac{1}{1-x} \end{align}$$

Answer 3

(Just playing around here)

Since $n^2 =\sum_{i=1}^n (2i-1) $,

$\begin{array}\\ \sum_{n=1}^{\infty} n^2x^n &=\sum_{n=1}^{\infty} x^n\sum_{i=1}^n (2i-1)\\ &=\sum_{n=1}^{\infty}\sum_{i=1}^n x^n (2i-1)\\ &=\sum_{i=1}^{\infty}\sum_{n=i}^{\infty} x^n (2i-1)\\ &=\sum_{i=1}^{\infty}(2i-1)\sum_{n=i}^{\infty} x^n \\ &=\sum_{i=1}^{\infty}(2i-1)\frac{x^i}{1-x} \\ &=\frac1{1-x}\sum_{i=1}^{\infty}(2i-1)x^i \\ &=\frac1{1-x}\left(2\sum_{i=1}^{\infty}ix^i-\sum_{i=1}^{\infty}x^i\right) \\ \end{array} $

Doing the same, but with $n =\sum_{i=1}^n 1 $,

$\begin{array}\\ \sum_{n=1}^{\infty} nx^n &=\sum_{n=1}^{\infty} x^n\sum_{i=1}^n 1\\ &=\sum_{n=1}^{\infty}\sum_{i=1}^n x^n \\ &=\sum_{i=1}^{\infty}\sum_{n=i}^{\infty} x^n \\ &=\sum_{i=1}^{\infty}\sum_{n=i}^{\infty} x^n \\ &=\sum_{i=1}^{\infty}\frac{x^i}{1-x} \\ &=\frac1{1-x}\sum_{i=1}^{\infty}x^i \\ &=\frac1{1-x}\frac{x}{1-x} \\ &=\frac{x}{(1-x)^2} \\ \end{array} $

Therefore (what do you know - looks like it's going to work!)

$\begin{array}\\ \sum_{n=1}^{\infty} n^2x^n &=\frac1{1-x}\left(2\sum_{i=1}^{\infty}ix^i-\sum_{i=1}^{\infty}x^i\right) \\ &=\frac1{1-x}\left(2\frac{x}{(1-x)^2} -\frac{x}{1-x}\right) \\ &=\frac1{(1-x)^3}\left(2x -x(1-x)\right) \\ &=\frac{x+x^2}{(1-x)^3} \\ \text{or}&=\frac{2x}{(1-x)^3}-\frac{x}{(1-x)^2} \\ \end{array} $