Power series where radius of convergence > lower limit

[Risborg]

Homework Statement

Let $\sum^{\infty}_{n=0} a_n(z-a)^n$ be a real or complex power series and set $\alpha = \limsup\limits_{n\rightarrow\infty} |a_n|^{\frac{1}{n}}$. If $\alpha = \infty$ then the convergence radius $R=0$, else $R$ is given by $R = \frac{1}{\alpha}$, where $0<R\leq\infty$.
A lower bound for the convergence radius can by found by using $\beta = \limsup\limits_{n\rightarrow\infty} \frac{|a_{n+1}|}{|a_n|}$, such that $\frac{1}{\beta} = \tilde{R}$, so $\tilde{R} \leq R$

Construct an example of a power series where $\tilde{R} \neq R$

The Attempt at a Solution

I have tried some different kinds of values for $a_n$, but I always end up with same answer for $R$ and $\tilde{R}$.
I think i need to define $a_n$ recursively, but I don't know how to prove it, below I've written the relation between $a_n$ and $a_{n+1}$.

$\beta^{-1} = \limsup\limits_{n\rightarrow\infty}\frac{|a_{n}|}{|a_{n+1}|} < \alpha^{-1} = \limsup\limits_{n\rightarrow\infty} |a_n|^{-\frac{1}{n}}$
which implies that
$\limsup\limits_{n\rightarrow0\infty} |a_{n+1}| > \limsup\limits_{n\rightarrow\infty} |a_{n}||a_n|^{\frac{1}{n}}$
I tried a few recursively defined $a_n$ but then I ended up with complicated expressions that I could find the limit of.

Can you think of a clever way to define $a_n$, or do you have some suggestions for what I could do next?

 

[Mark44]

Homework Statement

Let $\sum^{\infty}_{n=0} a_n(z-a)^n$ be a real or complex power series and set $\alpha = \limsup\limits_{x\rightarrow0} |a_n|^{\frac{1}{n}}$. If $\alpha = \infty$ then the convergence radius $R=0$, else $R$ is given by $R = \frac{1}{\alpha}$, where $0<R\leq\infty$.
A lower bound for the convergence radius can by found by using $\beta = \limsup\limits_{x\rightarrow0} \frac{|a_{n+1}|}{|a_n|}$, such that $\frac{1}{\beta} = \tilde{R}$, so $\tilde{R} \leq R$

Both of your limits are as x → 0, but x is not in the expressions you're taking the limits of. What should it be?

Construct an example of a power series where $\tilde{R} \neq R$

The Attempt at a Solution

I have tried some different kinds of values for $a_n$, but I always end up with same answer for $R$ and $\tilde{R}$.
I think i need to define $a_n$ recursively, but I don't know how to prove it, below I've written the relation between $a_n$ and $a_{n+1}$.

$\beta^{-1} = \limsup\limits_{x\rightarrow0}\frac{|a_{n}|}{|a_{n+1}|} < \alpha^{-1} = \limsup\limits_{x\rightarrow0} |a_n|^{-\frac{1}{n}}$
which implies that
$\limsup\limits_{x\rightarrow0} |a_{n+1}| > \limsup\limits_{x\rightarrow0} |a_{n}||a_n|^{\frac{1}{n}}$
I tried a few recursively defined $a_n$ but then I ended up with complicated expressions that I could find the limit of.

Can you think of a clever way to define $a_n$, or do you have some suggestions for what I could do next?

 

[Risborg]

Oops I forgot to change the variables when I copied from a latex example, it should be fixed now.

 

[mfb]

$\beta^{-1} = \limsup\limits_{n\rightarrow\infty}\frac{|a_{n}|}{|a_{n+1}|} < \alpha^{-1} = \limsup\limits_{n\rightarrow\infty} |a_n|^{-\frac{1}{n}}$
which implies that
$\limsup\limits_{n\rightarrow0\infty} |a_{n+1}| > \limsup\limits_{n\rightarrow\infty} |a_{n}||a_n|^{\frac{1}{n}}$

You cannot multiply like that.

You don't have to prove anything for the general case, it is sufficient to construct a counterexample. You want β to be large and α to be small. How can you make the limsup of the ratios large without making all the elements large?

 

[Risborg]

Okay thank you so much for the help.
I think I found a solution by choosing $a_n = 2$ if $n$ is even and $a_n = 3$ if $n$ is odd, then I end up having $\tilde{R} = \frac{2}{3} < 1 = R$.

 

[mfb]

Correct.