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Thread: 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 $\displaystyle \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$.

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


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

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

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

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

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

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

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

    Now, if $\displaystyle x \geqslant 1$, then

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

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

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

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


    Any ideas how to prove $\displaystyle \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$ imitating the above proof and considering $\displaystyle n^{\frac{1}{2n}}$? 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 $\displaystyle \lim_{n\rightarrow \infty} n^{\frac{1}{n}}=1$.

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


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

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

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

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

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

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

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

    Now, if $\displaystyle x \geqslant 1$, then

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

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

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

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


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

    hence: $\displaystyle 1 \leq n^{\frac{1}{n}} \leq \left(1 + \frac{\sqrt{n}-1}{n} \right)^2.$ 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; Oct 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 $\displaystyle a_{n} = \sqrt{n}$ to solve it.
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