# Finding a power series representation for $\frac{1}{(2-x)^2}$ in powers of $x$

Problem: Determine a power series representation for the function \begin{align*} \frac{1}{(2-x)^2} \end{align*} in powers of $x$. On what interval is the representation valid?

Attempt: We have \begin{align*} \frac{1}{2-x} = \frac{1}{2} \frac{1}{(1- \frac{x}{2})} = \sum_{n=0}^{\infty}\frac{1}{2} \bigg( \frac{x}{2} \bigg)^n \end{align*} So now I differentiated this last series, which gave me \begin{align*} \sum_{n=1}^{\infty} \frac{n}{2} \bigg( \frac{x}{2} \bigg)^{n-1} = \frac{1}{(2-x)^2} \end{align*} for $-2 < x < 2$. I then rewrote this as \begin{align*} \sum_{n=0}^{\infty} \frac{(n+1) x^n}{2^{n+1}} \end{align*} But my calculus textbook says it should be \begin{align*} \sum_{n=0}^{\infty} \frac{n+1}{2^{n+2}} x^n. \end{align*} Why the extra factor $2$ in the denumerator? Did I overlook something?

• Chain rule, $\dfrac{d}{dx}\biggl(\dfrac{x}{2}\biggr)^n = n\biggl(\dfrac{x}{2}\biggr)^{n-1}\cdot \dfrac{1}{2}$. Aug 16 '15 at 11:03

## 1 Answer

In differentiating you have forgeted a $\frac{1}{2}.$ The derivative of \begin{align*} \sum_{n=0}^{\infty} \frac{1}{2} \bigg( \frac{x}{2} \bigg)^{n} \end{align*} is \begin{align*} \sum_{n=1}^{\infty} \frac{n}{2} \bigg( \frac{x}{2} \bigg)^{n-1}.\frac{1}{2} \end{align*} Which you forgot the last $\frac{1}{2}$ and so the answer is is \begin{align*} \sum_{n=0}^{\infty} \frac{(n+1) x^n}{2^{n+2}}\end{align*}

• HI, nice work +1 Jan 17 at 13:44