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.