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Math Help - A confusing sequence question

  1. #1
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    Acolman, Mexico
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    A confusing sequence question

    Hello, I am completely lost with this question:

    Let x_n = n^{\frac{1}{n}} for n \in \mathbb{N}.
    1. Show that x_{n+1} < x_n if and only if (1+ \frac{1}{n}) < n and infer that the inequality is valid for n \geq 3. Conclude that (x_n) is ultimately decreasing and that x = \lim(x_n) exists.

    2. Use the fact that the subsequence (x_{2n}) also converges to x to conclude that  x=1.


    I see why (1+ \frac{1}{n}) < n given that n \geq 3 , since \lim (1+ \frac{1}{n}) = e < 3
    But that's pretty much what I got =(
    Last edited by akolman; August 17th 2010 at 06:39 PM.
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  2. #2
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    Since you wrote " \lim (1+ \frac{1}{n}) = e < 3", I'm going to assume that you mean (1+\frac{1}{n})^n instead.

    We have the following bidirectional causal chain:

    \displaystyle \Leftrightarrow  x_{n+1} > x_n
    \displaystyle \Leftrightarrow  n^{1/n} > (n+1)^{\frac{1}{n+1}}
    \displaystyle \Leftrightarrow  n^{\frac{n+1}{n}} > n+1
    \displaystyle \Leftrightarrow  n^{1+\frac{1}{n}} > n+1
    \displaystyle \Leftrightarrow  n.n^{\frac{1}{n}} > n+1
    \displaystyle \Leftrightarrow  n^{\frac{1}{n}} > \frac{n+1}{n} = 1+\frac{1}{n}
    \displaystyle \Leftrightarrow  n > \left(1+\frac{1}{n}\right)^n

    You should be able to do the rest now!
    Last edited by Vlasev; August 17th 2010 at 09:33 PM.
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