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…
Best Answer
If you try to use Cauchy-Hadamard theorem, you would get Radius too.
So as you mentoined let $a_n = 3^{-\frac{n(n+1)}{2}} = 3^{-\frac{n^2}{2}}3^{-n}3^{-\frac{1}{2}}$ . So radius is now: $$ 1/R = \limsup_{n \to \infty} \big|a_n\big|^{\frac{1}{n^2}} = \limsup_{n \to \infty} \Big| 3^{-\frac{n^2}{2}}3^{-n}3^{-\frac{1}{2}} \Big|^{\frac{1}{n^2}} = 3^{-\frac{1}{2}} $$
So the radius of convergence should be: $ R = \sqrt{3} $