Finding interval of convergence for series

Question

Find the interval of convergence and radius of convergence for the series:

$$ \sum_{n= 0}^\infty \frac{x^n}{3^n} $$

I'm not sure if I'm correct, but would the interval of convergence be $(-3,3)$ (not inclusive)
and would R = 3?

Answer 1

This is the geometric series with parameter $$p=\frac{x}{3}$$ and converges for values of $|p|<1$. So your answer is correct.

Answer 2

$$\lim_{n\to\infty}\left|\dfrac{x^{n+1}}{3^{n+1}}\dfrac{3^n}{x^n}\right|<1\\ \implies \lim_{n\to\infty}\left|\dfrac{x}{3}\right|<1\\ \implies |x|<3,\text{ so the sum converges when }x\in(-3,3),\text{ and the radius of convergence is therefore }3.$$

Answer 3

$$r=\lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|$$ if the limit exists. In your case you immediately get $r=3$.