Radius of convergence of the power series $\sum_{n=1}^{\infty}a_nz^{n^2}$ 
Find the radius of convergence of the power series $$\sum_{n=1}^{\infty}a_nz^{n^2}$$ where , $a_0=1$ and $a_n=\frac{a_{n-1}}{3^n}$.

My Work :
We, have, $$\frac{a_1}{a_0}.\frac{a_2}{a_1}...\frac{a_n}{a_{n-1}}=\frac{1}{3}.\frac{1}{3^2}...\frac{1}{3^n}$$
$$\implies a_n=3^{-\frac{n(n+1)}{2}}$$
Now put $n^2=p$. Then the series becomes , $$\sum_{p=1}^{\infty}a_pz^p$$ where, $a_p=3^{\frac{p+\sqrt {p}}{2}}$.
Now $$\left|\frac{a_{p+1}}{a_p}\right|\to \frac{1}{\sqrt 3} \text{ as } p\to \infty$$
So , radius of convergence is $\sqrt 3$.
I think my answer is correct..But I want to know does there any other trick so that we can find the radius of convergence without finding $a_n$ explicitly ?
Please help...

Just to add, if $\sum_n b_nz^n$ is a power series, then there is an explicit formula for the radius of convergence: $$R = \frac{1}{\limsup_n |b_n|^{\frac{1}{n}}}$$ In this case we have that $b_{n^2}=a_n=3^{-\frac{n(n+1)}{2}}$. Hence $b_{n^2}^{\frac{1}{n^2}}=3^{-\frac{1}{2}(1+\frac{1}{n})}$. Thus, substituting $k$ for $n^2$, we have that 
$|b_k|^{-\frac{1}{k}} = \left\{
 \begin{array}{}
    3^{-\frac{1}{2}(1+\frac{1}{\sqrt{k}})} & : k \text{ is a perfect square}\\
    0 & : \text{ otherwise}
  \end{array} \right.$
It follows that $\limsup_k |b_k|^{\frac{1}{k}}=3^{-\frac{1}{2}}$, which gives $R=\sqrt{3}$.
