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Sequence limit - real analysis

  1. Sep 27, 2007 #1
    1. The problem statement, all variables and given/known data

    Prove that the sequence [tex](x_n) = ((a^n+b^n)^{1/n})[/tex] converges to [tex]b[/tex], for 0 < a < b.

    3. The attempt at a solution

    I haven't dealt with any sequences with [tex]n[/tex]'s in the exponent, but I assume I'll have to use logarithms at some point to get at them? Can someone start me off in the right direction?
     
    Last edited: Sep 27, 2007
  2. jcsd
  3. Sep 27, 2007 #2

    Hurkyl

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    The whole idea is that a introduces an increasingly insignificant error. Therefore, differential calculus should be applicable. Personally, I would try a differential approximation.
     
    Last edited: Sep 27, 2007
  4. Sep 27, 2007 #3
    here's a follow-up attempt

    What about something like [tex]|(a^n+b^n)^{1/n}-b| < (2b^n)^{1/n}-b < 2^{1/n} < 2^{1/\epsilon}[/tex] when [tex]n > \epsilon[/tex]?
     
  5. Sep 27, 2007 #4

    Hurkyl

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    Why is [itex](2b^n)^{1/n} - b < 2^{1/n}[/itex]?

    And more importantly, why would [itex]|(a^n + b^n)^{1/n} - b| < 2^{1/\epsilon}[/itex] be helpful?
     
    Last edited: Sep 27, 2007
  6. Sep 27, 2007 #5
    i think i meant...

    [tex](2b^n)^{1/n}- b = 2^{1/n}b - b < 2^{1/n}b[/tex]

    sorry, i think it's just because it's late.
     
  7. Sep 27, 2007 #6

    Hurkyl

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    See the new question I added.
     
  8. Sep 27, 2007 #7
    if [tex]n>\epsilon[/tex], then [tex]\frac{1}{n} < \frac{1}{\epsilon}[/tex], and [tex]2^{\frac{1}{n}} < 2^{\frac{1}{\epsilon}} [/tex]
     
    Last edited: Sep 27, 2007
  9. Sep 27, 2007 #8

    VietDao29

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    Well, and how can you go from there? :rolleyes:

    Remember that:
    [tex]\lim_{\epsilon \rightarrow + \infty} 2 ^ {\frac{1}{\epsilon}} = 1[/tex], not 0.

    How about taking bn outside, like this:

    [tex]\lim_{n \rightarrow \infty} \left\{ b ^ n \left[ \left( \frac{a}{b} \right) ^ n + 1 \right] \right\} ^ \frac{1}{n}[/tex]

    Can you go from here? :)
     
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