# Radius of convergence without the series

Let $\displaystyle\sum_{n=2}^{\infty} a_nx^n$ be a power series, find the radius of convergence when

a)$\displaystyle \lim_{n\to \infty} \dfrac{a_n}{3^n} = 1$

I'm not sure what the question is asking, how does the limit of $a_n/3^n$ help me find the radius of convergence?

• What methods for determining the radius of convergence of a power series do you know? – Daniel Fischer Feb 26 '14 at 21:24
• @DanielFischer We haven't done any examples, but I know the root test and ratio test, as well as the definition of the radius of convergence as the supremum of a set $S = \{r\geq 0 | ... \}$ – Warz Feb 26 '14 at 21:26
• The root test and the ratio test both work with the given information. If you apply either of them to the given situation, what do you get? – Daniel Fischer Feb 26 '14 at 21:27
• @DanielFischer I don't understand what information we're given, how can I compute $\dfrac{|a_{n+1}x^{n+1}|}{|a_nx^n|}$ – Warz Feb 26 '14 at 21:29
• If $\left\lvert \frac{a_{n+1}}{a_n}\right\rvert \to \infty$, then we have for every $x\neq 0$ $$\left\lvert \frac{a_{n+1}x^{n+1}}{a_nx^n}\right\rvert = \left\lvert \frac{a_{n+1}}{a_n}\right\rvert\cdot \lvert x\rvert \to\infty,$$ so the series does not converge for $x$. – Daniel Fischer Feb 26 '14 at 22:09

$$\left\lvert\frac{a_{n+1}x^{n+1}}{a_nx^n}\right\rvert = \frac{3^{n+1}}{3^n}\left\lvert\frac{\frac{a_{n+1}}{3^{n+1}}x^{n+1}}{\frac{a_n}{3^n}x^n}\right\rvert.$$
$$\lim_{n\to\infty}\frac{a_n}{3^n} = 1.$$