Next, we know we need the denominator to be in the form \(1 - p\) and again that is easy enough, in this case, to rewrite the denominator by factoring a 3 out of the denominator as follows,

\[f\left( x \right) = \frac{{{x^3}}}{3}\frac{1}{{1 - \frac{1}{3}{x^2}}}\] Show Step 3

At this point we can use the formula from the notes to write this as a power series. Doing this gives,

\[f\left( x \right) = \frac{{{x^3}}}{3}\frac{1}{{1 - \frac{1}{3}{x^2}}} = \frac{{{x^3}}}{3}\sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{3}{x^2}} \right)}^n}} \hspace{0.25in}\hspace{0.25in}{\mbox{provided}}\,\,\left| {\frac{1}{3}{x^2}} \right| < 1\] Show Step 4

Now, recall the basic “rules” for the form of the series answer. We don’t want anything out in front of the series and we want a single \(x\) with a single exponent on it.

These are easy enough rules to take care of. All we need to do is move whatever is in front of the series to the inside of the series and use basic exponent rules to take care of the \(x\) “rule”. Doing all this gives,

\[f\left( x \right) = \frac{{{x^3}}}{3}\sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{3}{x^2}} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{1}{3}{x^3}{{\left( {\frac{1}{3}} \right)}^n}{{\left( {{x^2}} \right)}^n}} = \sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{3}} \right)}^{n + 1}}{x^{2n + 3}}} \hspace{0.25in}\hspace{0.25in}\,{\mbox{provided}}\,\,\left| {\frac{1}{3}{x^2}} \right| < 1\] Show Step 5

To get the interval of convergence all we need to do is do a little work on the “provided” portion of the result from the last step to get,

\[\left| {\frac{1}{3}{x^2}} \right| < 1\,\,\,\, \to \,\,\,\,\,\frac{1}{3}{\left| x \right|^2} < 1\hspace{0.25in} \to \hspace{0.25in}{\left| x \right|^2} < 3\hspace{0.25in} \to \hspace{0.25in}\left| x \right| < \sqrt 3 \hspace{0.25in} \to \hspace{0.25in} - \sqrt 3 \, < x < \sqrt 3 \]

Note that we don’t need to check the endpoints of this interval since we already know that we only get convergence with the strict inequalities and we will get divergence for everything else.

Show Step 6

The answers for this problem are then,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Power Series : }}\frac{{{x^3}}}{{3 - {x^2}}} = \sum\limits_{n = 0}^\infty {{{\left( {\frac{1}{3}} \right)}^{n + 1}}{x^{2n + 3}}} }}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Interval : }} - \sqrt 3 \, < x < \sqrt 3 \,}}\]