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Homework Help: Root test question

  1. Jul 21, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider the following series (Baby Rudin Example 3.35):
    [itex]\frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \ldots[/itex]

    We have to apply the root test.

    2. The attempt at a solution
    [itex]\sqrt[n]{2^{-\frac{n+1}{2}}} = \sqrt[2n]{2^{-(n+1)}}[/itex] for n odd; (1)
    [itex]\sqrt[n]{3^{-\frac{n}{2}}} = \sqrt[2n]{3^{-n}}[/itex] for n even.

    Letting [itex]n\to\infty[/itex] we have [itex]1/\sqrt{3}[/itex] and [itex]1/\sqrt{2}[/itex] for inferior and superior limits. This result is equivalent to that one of Rudin, however, he states, that for n odd, we have [itex]\sqrt[2n]{2^{-n}}[/itex] in Eq. (1).

    What is the obvious thing that I'm still missing? Thanks for responses in advance!
  2. jcsd
  3. Jul 21, 2011 #2
    That's not what Rudin says. The only thing that Rudin says is


    which is certainly true. I think you may be confused because he left out a small step

    [tex]\limsup{\sqrt[n]{a_n}} = \lim{\sqrt[2n]{2^{-(n+1)}}} = \lim{\sqrt[2n]{2^{-1}}\sqrt[2n]{2^{-n}}}=\lim{\sqrt[2n]{2^{-n}}}[/tex]
  4. Jul 22, 2011 #3
    I don't have the Rudin text available to me.

    However, can't we bound the series above to establish convergence by
    [itex]\sum(1/2^k + 1/2^k)[/itex] from k = 1 to oo?
    If necessary, find a way to reindex the given series to match.
    The bounding series then becomes,
    [itex]\sum(2 * 1/2^k)[/itex] from k = 1 to oo,

    Maybe that bypasses, what we're supposed to be learning.
  5. Jul 22, 2011 #4
    Thank you, I was confused with this. I had the same idea, but I didn't see the reason for solving the limit for only one part of the formula. Of course, the equality of RHS and LHS is obvious. But that is the pure math and the Rudin's (correct) way :-)
  6. Jul 22, 2011 #5
    nickalh: of course, there are several possible ways how to solve it, but I was confused by the Rudin's way.
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