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

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.

• Show manually that the series diverges of $|\omega|>1$, and note it clearly converges if $|\omega|<1$. Hence the radius of convergence is $1$. – Pedro Tamaroff Nov 6 '16 at 0:18
• What do you mean by "convert" to a power series? It is a power series already - with many of the coefficients equal to 0. – mathguy Nov 6 '16 at 0:50
• I tried to found an explicit power series where de zero coefficients appears, and then use the root test. – bpittcher Nov 6 '16 at 1:07
• One could also differentiate with respect to $w$ and find that series' radius of convergence. – Simply Beautiful Art Nov 6 '16 at 1:10

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.