Radius of convergence of $\sum_{n=0}^{\infty} \frac{w^{2^n}}{2^n}$

Question

I am trying to solve this problem, but I'm unable to make progress.

I need to obtain the radius of convergence of $\displaystyle\sum_{n=0}^{\infty} \dfrac{w^{2^n}}{2^n}$.

I tried to convert this to a power series in order to apply the radius of convergence criterion, but i could not.

Answer 1

By the Cauchy-Hadamard theorem $$ \frac{1}{R}=\limsup_{n\to\infty}\left(\frac{1}{2^n}\right)^{1/2^n}=\limsup_ne^{\frac{-n\log2}{2^n}}=e^0=1, $$ where $R$ is the radius of convergence.