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?
 A: $$\sum_{n=0}^{+\infty}\frac{x^{4n}}{9^{n+1}} = \sum_{n = 0}^\infty \frac 19\cdot\left(\frac{x^4}{9}\right)^n$$
The nth root gives us $$x^4 \lt 9 \implies |x| \lt \sqrt 3$$
A: 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 $4$.
A: 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.
A: $$\sum_{n=0}^{+\infty}\frac{x^{4n}}{9^{n+1}}$$
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.
