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Math Help - Need to show that lim[n^(1/n)]=1. Please help!

  1. #1
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    Unhappy Need to show that lim[n^(1/n)]=1. Please help!

    Hi! Can anyone please help me with the following proof:

    Show that \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1.

    I have to show this using the AM/GM inequality, imitating the proof that \lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1 for x>0 and considering n^{\frac{1}{2n}}.


    This is how I proved that \lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1 for x>0 :

    The Arithmetic/Geometric mean inequality states that, given non-negative a_{1},  a_{2},  a_{3}, ...  a_{n},

    GM = (a_{1} a_{2} ... a_{n})^\frac{1}{n} \leq  \frac{a_{1} + a_{2} + ... + a_{n}}{n} = AM.

    Now I apply the AM/GM inequality to the numbers a_{1} = a_{2} = ... = a_{n-1} = 1 and a_{n} = x.

    Then G_{n} = x^{\frac{1}{n}} and A_{n} = \frac{(n - 1 + x)}{n} = \frac{(x - 1)}{n} + 1.

    Hence x^\frac{1}{n} \leq \frac{x - 1}{n} + 1

    i.e. x^\frac{1}{n} - 1 \leq \frac{x - 1}{n}.

    Now, if x \geqslant 1, then

    0 \leq x^\frac{1}{n} -1 \leq \frac{x - 1}{n}

    and hence x^\frac{1}{n} \rightarrow 1 as n \rightarrow \infty by the sandwich theorem.

    If 0 < x < 1, then x = y^{-1} where y > 1. But y^\frac{1}{n} \rightarrow 1 as n \rightarrow \infty and hence

    x^\frac{1}{n} = \frac{1}{y^\frac{1}{n}} \rightarrow \frac{1}{1} = 1 as n \rightarrow \infty by the combination theorem.


    Any ideas how to prove \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1 imitating the above proof and considering n^{\frac{1}{2n}}? I'd appreciate any help!
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  2. #2
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    Quote Originally Posted by clubbed2death View Post
    Hi! Can anyone please help me with the following proof:

    Show that \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1.

    I have to show this using the AM/GM inequality, imitating the proof that \lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1 for x>0 and considering n^{\frac{1}{2n}}.


    This is how I proved that \lim_{n\rightarrow \infty} x^{\frac{1}{n}}=1 for x>0 :

    The Arithmetic/Geometric mean inequality states that, given non-negative a_{1}, a_{2}, a_{3}, ... a_{n},

    GM = (a_{1} a_{2} ... a_{n})^\frac{1}{n} \leq \frac{a_{1} + a_{2} + ... + a_{n}}{n} = AM.

    Now I apply the AM/GM inequality to the numbers a_{1} = a_{2} = ... = a_{n-1} = 1 and a_{n} = x.

    Then G_{n} = x^{\frac{1}{n}} and A_{n} = \frac{(n - 1 + x)}{n} = \frac{(x - 1)}{n} + 1.

    Hence x^\frac{1}{n} \leq \frac{x - 1}{n} + 1

    i.e. x^\frac{1}{n} - 1 \leq \frac{x - 1}{n}.

    Now, if x \geqslant 1, then

    0 \leq x^\frac{1}{n} -1 \leq \frac{x - 1}{n}

    and hence x^\frac{1}{n} \rightarrow 1 as n \rightarrow \infty by the sandwich theorem.

    If 0 < x < 1, then x = y^{-1} where y > 1. But y^\frac{1}{n} \rightarrow 1 as n \rightarrow \infty and hence

    x^\frac{1}{n} = \frac{1}{y^\frac{1}{n}} \rightarrow \frac{1}{1} = 1 as n \rightarrow \infty by the combination theorem.


    Any ideas how to prove \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1 imitating the above proof and considering n^{\frac{1}{2n}}? I'd appreciate any help!
    let n \in \mathbb{N}. put a_1 = \cdots = a_{n-1}=1, \ a_n = \sqrt{n}. then by AM-GM: (\sqrt{n})^{\frac{1}{n}} \leq \frac{n -1+\sqrt{n}}{n}=1 + \frac{\sqrt{n} - 1}{n}. thus: n^{\frac{1}{n}} \leq \left(1 + \frac{\sqrt{n}-1}{n} \right)^2.

    hence: 1 \leq n^{\frac{1}{n}} \leq \left(1 + \frac{\sqrt{n}-1}{n} \right)^2. now apply the squeeze theorem to finish the proof.
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  3. #3
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    Look here.

    EDIT: Bad post. Cannot delete since already been responded to.
    Last edited by ThePerfectHacker; October 26th 2008 at 08:56 PM.
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  4. #4
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    Cheers NonCommAlg ! Can't believe I didn't think of using a_{n} = \sqrt{n} to solve it.
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