Dtyjlui

So I need to find the Convergence Radius and Convergence Domain of the complex power series:
$sumlimits_{n=0}^infty frac{1}{2^{n}} z^{2^n}$

I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.

It can probably be solved, after a step or two, by the formula $R=frac{1}{displaystylelimsup_{nrightarrow infty}sqrt[n]{|a_n|}}$ .

Thanks.

For $n ge 0$, let $a_n = begin{cases} frac1n, & n = 2^ktext{ for } k in mathbb{N}\ 0,&text{otherwise}end{cases}$.

The radius of convergence $R$ of the power series
$$sum_{k=0}^infty frac{1}{2^k} z^{2^k} = sum_{n=1}^infty a_n z^n$$
can be computed using the root test.

$$begin{array}{rll}
frac1R &= limsup_{ntoinfty} |a_n|^{1/n}
& color{blue}{leftarrow text{root text}}\
&= lim_{ntoinfty} sup_{mge n} |a_m|^{1/m}
& color{blue}{leftarrow text{definition of "limsup"}}\
&= lim_{ntoinfty}sup_{2^kge n} 2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{only $m$ of the from $2^k$ matter}}\
&= lim_{ntoinfty}sup_{k ge lceil log_2nrceil} 2^{-frac{k}{2^k}}\
&= lim_{elltoinfty}sup_{k ge ell}2^{-frac{k}{2^k}}
& color{blue}{leftarrow text{ $ell = lceil log_2 n rceil to infty$ as $n to infty$}}\
&= limsup_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrow text{definition of "limsup" again}}\
&= lim_{elltoinfty} 2^{-frac{ell}{2^ell}}
& color{blue}{leftarrowtext{ limsup = lim whenvever limit exists}} \
&= 2^{-lim_{elltoinfty} frac{ell}{2^ell}}
& color{blue}{leftarrow 2^{-x} text{ is a continuous function in } x}
\
&= 2^0 = 1
end{array}
$$

The radius of convergence of is $1$. The power series converges to some function $f(z)$ analytic over the open unit disk $|z| < 1$.

For any point $z$ on the unit circle, we have

$$left| sum_{nto 0}^infty a_n z^n right| le sum_{n=0}^infty |a_n z^n|
= sum_{nto 0}^infty |a_n| = sum_{k=0}^infty frac{1}{2^k} = 1$$

The power series converges absolutely and hence converges over the unit circle. This means the power series converges over the whole closed unit disk $|z| le 1$.

Notice the indices where $a_n ne 0$ is $n = 2^k$ and $frac{2^k}{k}$ diverges to $infty$ as $k to infty$. By Fabry gap theorem, the unit circle is a natural boundary for the function $f$.
There is no way to analytic continue $f(z)$ outside the closed unit disk.

This means the domain of the power series is exactly the closed unit disk (even if one
allow analytic continuation).

Using the Cauchy Condensation test, which states that the series of the sequence $f(n)$ converges if and only if the series of the sequence $2^n f(2^n)$ converges, the convergence of $a_n=z^{2^n}/2^n$ is equivalent to the convergence of $b_n=z^n/n^2$. Applying the root test and checking the boundary case shows that the series converges whenever $|z|le 1$.

i) If $|z|<1,$ then $|z^{2^n}/2^n| le 2^{-n}.$ Since $sum 2^{-n}<infty,$ the radius of convergence of the series is at least $1.$ ii) If $|z|>1,$ then $|z|^m/mto infty$ as $mto infty.$ Thus $|z^{2^n}/2^n|to infty.$ It follows that the series diverges for such $z,$ which implies the radius of convergence is at most $1.$ Putting i) and ii) together shows the radius of convergence is $1.$