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Power series where radius of convergence > lower limit

  1. Sep 25, 2015 #1
    1. The problem statement, all variables and given/known data
    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##

    3. 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?
     
    Last edited: Sep 25, 2015
  2. jcsd
  3. Sep 25, 2015 #2

    Mark44

    Staff: Mentor

    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?

     
  4. Sep 25, 2015 #3
    Oops I forgot to change the variables when I copied from a latex example, it should be fixed now.
     
  5. Sep 25, 2015 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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?
     
  6. Sep 26, 2015 #5
    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##.
     
  7. Sep 26, 2015 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Correct.
     
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