Finding the Radius of Convergence of the complex power series

Question

The power series I have is this:

$$ \sum_{n=0}^\infty a_n z^n ~~~\text{such that}~~ \sum_{n=0}^\infty 2^n a_n ~~\text{converges while}~~\sum_{n=0}^\infty (-1)^n2^na_n~~\text{diverges}$$

This is my attempt:

let $u_n = 2^na_n $, then $\limsup \sqrt[n]{|u_n|} = \limsup\sqrt[n]{|2^na_n|} = 2 \limsup\sqrt[n]{|a_n|} = l_1 < 1 $

Also $\limsup\sqrt[n]{|(-1)^n2^na_n|} = 2 \limsup\sqrt[n]{|a_n|} = l_2 > 1$

Then the radius of convergence of $\sum a_n z^n$ is at least $\frac{2}{l_1}$ and at most $\frac{2}{l_2}$. So the radius is between these two values.

I'm not sure if my approach is right because I'm not getting a definite value.

Any help would be much appreciated.

Edit:

Is it true that $l_1 = l_2$? And therefore in this case the radius = $2$.

Answer 1

Your approach is correct.

You have got $$2 \lim \sup \sqrt[n] {|a_n|} \le 1 \Rightarrow 2 \le \frac {1}{\lim \sup \sqrt[n] {|a_n|}}=R........(1)$$

And $$2 \lim \sup \sqrt[n] {|a_n|} \ge 1 \Rightarrow 2 \ge \frac {1}{\lim \sup \sqrt[n] {|a_n|}}=R........(2)$$

From $(1)$ and $(2)$, $R=2$.