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

Jun 22, 2018

$\frac{1}{1 - x} ^ 2 = 1 + 2 x + 3 {x}^{2} + \ldots$

#### Explanation:

We are given

$f \left(x\right) = \frac{1}{1 - x} ^ 2$

This is fairly similar to $\frac{1}{1 - x}$, for which we know a power series:

$\frac{1}{1 - x} = 1 + x + {x}^{2} + \ldots = {\sum}_{k = 0}^{\infty} {x}^{k}$

The radius of convergence for this power series is $x \in \left(- 1 , 1\right)$.

While it would be easy to say that

$\frac{1}{1 - x} ^ 2 = {\left({\sum}_{k = 0}^{\infty} {x}^{k}\right)}^{2}$

This is not a valid representation of a power series.

Usually, some power series arise from derivatives. It'd be worth a shot to try this one, too.

"d"/("d"x) [1/(1-x)] = "d"/("d"x) [1+x+x^2+...]

By the quotient rule,

"d"/("d"x) [1/(1-x)] = - ("d"/("d"x) [1-x])/(1-x)^2=color(red)(1/(1-x)^2

As "d"/("d"x) x^k = kx^(k-1):

"d"/("d"x) [1+x+x^2+...] = 0 + 1 + 2x + 3x^2 + ... = sum_(k=0)^oo kx^(k-1)

Hence the power series representation of $f \left(x\right)$ is

$\frac{1}{1 - x} ^ 2 = {\sum}_{k = 0}^{\infty} k {x}^{k - 1}$

with radius of convergence $x \in \left(- 1 , 1\right)$.