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The power series I have is this:

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

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

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    Why not just let $a_n=\frac{(-1)^n}{n2^n}$? – 2017-02-19
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    but I am allowed to take a specific value for $a_n$..am I wrong? – 2017-02-19
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    If you are doing a problem that asks you to find an example of such a sequence, then sure, you are allowed your own choices. How could it be otherwise? There's an infinity of possible solutions out there! And if that is not the problem statement, well, you haven't told us what is being asked. – 2017-02-19
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    The $<1$ and $>1$ claims are incorrect. – 2017-02-19

1 Answers 1

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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 $(2)$, $R=2$.