Radius of convergence and sum of the series

I have the series $$\sum_{n=0}^{+\infty}\frac{x^{4n}}{9^{n+1}}$$

I'm supposed to find the radius of convergence and sum this order. I have tried finding the radius by using $$R =\frac{1}{\overline\lim_{n \to +\infty}\left|\sqrt[n]{a_n}\right|}$$

Which gives me 9, but WolframAlpha says that the radius of convergence is $\{x : |x| < \sqrt{3}\}$. What am I doing wrong? And how am I supposed to sum this order?

• Your formula is valid for power series of the type $\sum{a_n x^{\color{red}{n}}}$ and not $\sum{a_n x^{\color{blue}{4}\color{red}{n}}}$ – jdoicj Jun 13 '14 at 11:47
• Using the root test is overkill in this case, since the series is geometric. – Michael Hardy Jun 13 '14 at 12:13

Let $$S = \sum_{n=0} \frac{x^{4n}} {9^{n+1}}.$$

Now look at $9S$: $$9S = \sum_{n=0} \frac{x^{4n}} {9^{n}}.$$ That looks easier to work with, right? Now replace $x^4$ with $y$ to get $$9S = \sum_{n=0} \frac{y^{n}} {9^{n}}.$$ NOW apply that to find the radius of convergence, $R_y$, for $y$, and take its fourth root to find the radius of convergence for $x$, etc.

• I understand now. Thank you. One more question though. Won't the series converge for $x = \pm \sqrt{3}$ as well, since the series will be $\sum_{n=0}^{+\infty}\frac{9}{9^{n+1}} = \sum_{n=0}^{+\infty}\frac{1}{9^n}$, which converges as it is a geometric series with $|q| = \frac{1}{9} < 1$? – Proka Jun 13 '14 at 12:15
• No. When you plug in $x = \sqrt{3}$, the numerator becomes $9^n$, not just $9$. :( – John Hughes Jun 13 '14 at 17:39

$$\sum_{n=0}^{+\infty}\frac{x^{4n}}{9^{n+1}} = \sum_{n = 0}^\infty \frac 19\cdot\left(\frac{x^4}{9}\right)^n$$

The nth root gives us $$x^4 \lt 9 \implies |x| \lt \sqrt 3$$

The value $a_n$ is the number that sits next to $x^n$. This means that $a_n$ is not $\frac{1}{9^{n+1}}$, but is $0$ for $n$ not divisible by $4$ and is $$\frac1{9^{\frac{n}{4} + 1}}$$ for $n$ which is divisible by $4$.

$$\sum_{n=0}^{+\infty}\frac{x^{4n}}{9^{n+1}}$$

This series is a geometric series. Those converge if and only if the absolute value of the common ratio is less than $1$. In this case the common ratio is $x^4/9$: every time $n$ increases by $1$, the term is multiplied by $x^4/9$. $$\frac{x^4}{9} <1$$ $$x^4<9$$ $$|x|<\sqrt[4]{9} = \sqrt{3}.$$ $$-\sqrt{3} < x < \sqrt{3}.$$ So the radius of convergence is $\sqrt{3}$.

Wolfram alpha may be giving you the set of values of $x$ for which the series converges, but the radius of convergence is a number, not a set.