My aim is to find the radius of convergence of the series $\sum a_n x^{n^2}$ such that $a_0=1$ and $a_n=3^{-n}a_{n-1}$ for all $n\in\mathbb N$.
I will use the Ratio test, let $c_n=a_nx^{n^2}$
$$\lim \frac{c_n}{c_{n-1}} = \lim \frac{a_n\cdot x^{n^2}}{a_{n-1}\cdot x^{(n-1)^2}} = \lim \frac{x^{2n-1}}{3^{n}}$$
Now to find $\lim \frac{x^{2n-1}}{3^{n}}$, I will use the Root test
$$\lim |\frac{x^{2n-1}}{3^{n}}|^{\frac1n}=\frac{x^2}{3} $$
if $\frac{x^2}{3}<1 $ or $|x|<\sqrt{3}$ $ \;\Rightarrow \frac{x^{2n-1}}{3^{n}}\to 0 $
and if if $\frac{x^2}{3}>1 $ or $|x|>\sqrt{3} \Rightarrow \frac{x^{2n-1}}{3^{n}} \to \infty$
Since from the Ratio test, we need $\lim \frac{x^{2n-1}}{3^{n}}<1$ Hence the given series will converge if $|x|<\sqrt{3}$. So radius of convergence must be $R=\sqrt{3}$. However, my professor told us that the radius of convergence will be $\infty$. Can someone please tell me which is the correct answer and where did I go wrong in my explanation?