What would be the radius of convergence of $\sum\limits_{n=0}^{\infty} z^{3^n}$? I know how to find the radius of convergence of a power series $\sum\limits_{n=0}^{\infty} a_nz^n$, but how does this apply to the power series $\sum\limits_{n=0}^{\infty} z^{3^n}$? Would the coefficients $a_n=1$, so that one may apply D'Alembert's ratio test to determine the radius of convergence? I would appreciate any input that would be helpful.

What do you do for the cosine and sine series? There, you cannot use the Ratio Test directly because every other coefficient is equal to $0$. Instead, we do the Ratio Test on the subsequence of even (resp. odd) terms. You can do the same here. We have $a_{3^n}=1$ for all $n$, and $a_j=0$ for $j$ not a power of $3$. Define $b_k = z^{3^k}$. Then we are trying to determine the convergence of the series $\sum b_k$, so using the Ratio Test we have:
$$\lim_{n\to\infty}\frac{|b_{n+1}|}{|b_{n}|} = \lim_{n\to\infty}\frac{|z|^{3^{n+1}}}{|z|^{3^{n}}} = \lim_{n\to\infty}|z|^{3^{n+1}-3^n} = \lim_{n\to\infty}|z|^{2\times3^{n}} = \left\{\begin{array}{cc}
0 & \text{if }|z|\lt 1\\
1 & \text{if }|z|=1\\
\infty &\text{if }|z|\gt 1
\end{array}\right.$$
So the radius of convergence is $1$. 
