Convergence of Power Series with $z^{2^k}$ So i got this  power series $$\sum_{k=0}^\infty 3^k \cdot z^{2^k}$$

Can i do the convergence test like this:

$$\text{ratio test} \Longrightarrow \frac{3^{k+1}\cdot z^{2^{k+1}}}{3^k \cdot z^{2^k}} = \frac{3 \cdot z^{2^{k+1}}}{z^{2^k}} = 3 \cdot z^2 \lt 1 \Longleftrightarrow |z| &lt; \frac{1}{\sqrt3}$$
If you want to see your series as a power series, observe that

$$\sum_{k=0}^\infty 3^k z^{2^k}=1\cdot z+3z^2+9z^4+\cdots+3^kz^{2^k}+\cdots$$

observe then that the sequence of coefficients is

$$a_n:=\begin{cases}0,&amp;n\neq 2^k\\{}\\3^k,&amp;n=2^k\end{cases}$$

Thus, the convergence radius is given by (the sequence above is non-negative):

$$\frac1R=\lim\sup_{n\to\infty}\sqrt[n]{a_n}=\lim_{n=2^k\to\infty\iff k\to\infty}3^{k/2^k}=1\implies R=1$$
The Ratio test gives the limit $ L=\lim_{k\to +\infty} 3|z|^{2^k}$. 


$|z|&lt;1\implies L=0\implies R\geq 1$
$|z|=1 $ and $ \sum 3^k $ diverges $\implies R\leq 1$


thus, the radius of convergence is $R=1$.