Finding the rational function the power series $\sum_{n=0}^\infty \frac{(-1)^n}{3^{n+1}}x^{n+2}$ converges to

Question

Given the power series, find the rational function equation that the power series converges to.

$$\sum_{n=0}^\infty \frac{(-1)^n}{3^{n+1}}x^{n+2}$$

So far I've figured out that the rational function $\frac{1}{3(1+x)}$ is equal to $\sum \frac{(-1)^n}{3}x^n$

Can someone please explain how I get from here to the final answer? Specifically, what do I need to do to the rational function to turn the $n$'s into $(n+1)$ or $(n+2)$ ?

Answer 1

You're right that it is easier to work with all terms raised to the same power. To that end, note that $$ \sum_{n=0}^{\infty}\frac{(-1)^n}{3^{n+1}}x^{n+2}=\frac{1}{3}{x^2}\sum_{n=0}^{\infty}\frac{(-1)^n}{3^n}x^{n}=\frac{1}{3}{x^2}\sum_{n=0}^{\infty}\left(-\frac{x}{3}\right)^n. $$ The sum is now a simple geometric series, summing to $1/(1+x/3)$, which should lead you to the final result.

Answer 2

HINT

Can you compare the following expression with the geometric series?

\begin{align*} \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n+2}}{3^{n+1}} & = \frac{x^{2}}{3}\sum_{n=0}^{\infty}\left(-\frac{x}{3}\right)^{n} \end{align*}