Express $\frac{1}{(1x^4)}$ as the sum of a power series and find the interval of convergence.

Solution

We have the following equation.

$\frac{1}{1-x}=1x{x}^2{x}^3$ cdots$={\displaystyle \underset{n=0}{\overset{}{}}}{x}^n,|x|<mn>1</mn>$

Replacing $x$ by $-x^4$ in the equation above, we have the following.

$\frac{1}{1x^4}=\frac{1}{1-(-x^4)}={\displaystyle \underset{n=0}{\overset{}{}}}({)}^n$

$={\displaystyle \underset{n=0}{\overset{}{}}}(-1{)}^n{x}^{4n}=1-x^4,-x^{12}{x}^{16}-$cdots

Because this is a geometric series, it converges when that is$,x^4<mtext></mtext><mi></mi>$ or Therefore the interval of convergence is the open interval $-($Of course, we could have determined the radius of convergence by applying the Ratio Test, but that much work is unnecessary here.$)$