# How do you find a power series representation for f(x) = (x) / ((1-x)^2) and what is the radius of convergence?

Sep 28, 2015

Recall that:

The power series for $\frac{1}{1 - x}$ is ${\sum}_{n = 0}^{N} {x}^{n}$.

What you can do is take the derivative of both sides:

$\frac{d}{\mathrm{dx}} \left[{\sum}_{n = 0}^{N} {x}^{n}\right] = \frac{d}{\mathrm{dx}} \left[\frac{1}{1 - x}\right] = \textcolor{g r e e n}{\frac{d}{\mathrm{dx}} \left[1 + x + {x}^{2} + \ldots\right]}$

When we take the derivative, note that the first term will disappear ($\frac{d}{\mathrm{dx}} \left[1\right] = 0$), so the first $n$ shifts from $n = 0$ to $n = 1$.

sum_(n=1)^N nx^(n-1) = 1/(1-x)^2 = color(green)(?)

We can shift it back to $n = 0$ for conventional purposes.

How you do that is consider that you are currently starting at $n = 1$. That means $n - 1 = 0$. To get the exponent to become $0$ using $n = 0$ below $\sum$, $n - 1$ becomes $n$.

When $n = 1$ below $\sum$, $n {x}^{n - 1}$ starts at $1 {x}^{0}$, so when $n = 0$ below $\sum$, $n {x}^{\text{stuff}}$ becomes $\left(n + 1\right) {x}^{\text{stuff}}$.

color(highlight)(sum_(n=0)^N (n+1)x^(n) = 1/(1-x)^2) = color(green)(?)

So, all we need to do is do it for the explicit series to get:

$\frac{1}{1 - x} ^ 2 = \frac{d}{\mathrm{dx}} \left[1 + x + {x}^{2} + \ldots\right]$

$= \textcolor{b l u e}{1 + 2 x + 3 {x}^{2} + 4 {x}^{3} + \ldots}$

The radius of convergence is based on the idea that the sum of magnitudes $| a | \ge 1$ will not converge if they are all positive.

$\frac{1}{1 - x} ^ 2 > 1$ when $| x | \ge 1$.

Therefore, the radius of converge is:

$\textcolor{b l u e}{x \in \left(- 1 , 1\right)}$

or you can write it as:

$\textcolor{b l u e}{| x | < 1}$