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Radius of convergence and 2^1/2

  1. Jan 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Suppose c_n is the digit in the nth place of the decimal expansion of 2^1/2. Prove that the radius of convergence of [itex]\sum{c_n x^n}[/itex] is equal to 1.


    2. Relevant equations



    3. The attempt at a solution

    What I want to show is that limsup |c_n|^1/n = 1. Clearly for any c_n, (c_n)^1/n ≥ 1. So we have sup (c_n)^1/n ≥ 1 for any n. It is also easy to see that sup (c_n)^1/n ≤ n^1/n for n large enough, as c_n is at most 9. So by the Sandwich lemma limsup |c_n|^1/n = 1.

    Correct?
     
  2. jcsd
  3. Jan 16, 2013 #2

    jbunniii

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    Why is this clearly true? Are you saying that the decimal expansion of [itex]\sqrt{2}[/itex] doesn't contain any zero digits?
    Yes, this part is true, because there are infinitely many nonzero [itex]c_n[/itex]'s. (Proof?)

    [Edit]: Be careful about the notation. I think you meant
    [tex]\sup \{(c_m)^{1/m} : m \geq n\} \geq 1 \textrm{ for any } n[/tex]
    and therefore
    [tex]\limsup (c_n)^{1/n} \geq 1[/tex]
    Yes, that looks OK, assuming you have already proved that [itex]\lim_{n\rightarrow \infty} n^{1/n} = 1[/itex].
     
    Last edited: Jan 16, 2013
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