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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.

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

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