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Interesting proof of sequence limit

  1. Feb 9, 2008 #1
    Prove that sequence [tex]x_n=\frac{n}{\sqrt[n]{n!}}[/tex] is increasing and [tex]\lim_{n\rightarrow\infty}{x_n} = e[/tex]

    My attempt:
    First, I try to prove [tex]\forall n\in N:\frac{x_{n+1}}{x_n}>1[/tex].
    [tex]\forall n\in :\frac{x_{n+1}}{x_n}=\frac{n+1}{n}\frac{\sqrt[n]{n!}}{\sqrt[n+1]{(n+1)!}}
    =\sqrt[n(n+1)]{\frac{(n+1)^{n^2}n!}{n^{n^2+n}}}[/tex] and this is dead end.

    Another attempt, we have [tex]x_1 > 0, x_2 > 0, ..., x_n>0:\sqrt[n]{x_1x_2..x_n}\leq\frac{x_1+x_2+...+x_n}{n}[/tex], where equality is then [tex]x_1=x_2=...=x_n[/tex]
    So, [tex]\forall n\in N: \sqrt[n]{n!}<\frac{(1+n)n}{2n}=\frac{n+1}{2}
    \Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}>\frac{2n}{1+n}=y_n[/tex]
    Sequence [tex](y_n)[/tex] is increasing, because [tex]\forall n\in N:\frac{y_{n+1}}{y_n}=\frac{2(n+1)}{n+2}\frac{1+n}{n+2}=\frac{n^2+2n+1}{n+2}=n+\frac{1}{n+2}>1[/tex]

    Can I say, [tex]x_n[/tex] is increasing, because [tex]y_n[/tex] is increasing and [tex]x_n > y_n, n \in N[/tex] ?

    Next, we have [tex]n!>\left({\frac{n}{e}}\right)^n, n \in N[/tex]. I will omit the proof of this inequality, it is proven very simply by using math. induction.

    So, we have [tex]\forall n\in N:n!>\left({\frac{n}{e}}\right)^n \Longrightarrow e>\frac{n}{\sqrt[n]{n!}}[/tex]

    Next, [tex]\forall n \in N: \sqrt[n]{n!} < \frac{(n+1)n}{2n} = \left(1 + \frac{1}{n}\right) \frac{n}{2}<\left(1 + \frac{1}{n}\right)n=\frac{1}{1-\frac{1}{n+1}}n \Longrightarrow x_n=\frac{n}{\sqrt[n]{n!}}>1-\frac{1}{n+1}>1-\frac{1}{n}>\left(1-\frac{1}{n}\right)^{n}=z_n[/tex]

    Finally, we have [tex]\forall n \in N: e> x_n=\frac{n}{\sqrt[n]{n!}}>z_n[/tex]
    and [tex]\lim_{n\rightarrow\infty}z_n=e \Longrightarrow \lim_{n\rightarrow\infty}x_n=e[/tex]


    Is it correct ? Any suggestions ? I am missing something here...
    Last edited: Feb 9, 2008
  2. jcsd
  3. Feb 9, 2008 #2
    And another related problem with one above: Prove sequence [tex](y_n), y_n=\frac{n}{\left(\sqrt[n]{n!}\right)^2}[/tex] is decreasing and [tex]\lim y_n = 0 [/tex].
    I see [tex]y_n=\frac{x_n^2}{n}[/tex] and [tex]\lim y_n = \frac{\lim x_n^2}{\lim_{n\rightarrow\infty}n}=\frac{e^2}{\lim_{\rightarrow\infty}n}=0.[/tex]

    But how to show sequence is decreasing ?
  4. Feb 9, 2008 #3


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    No you can't.
    Also, you missed the ! in the original equation for x_n.

    The first thing I see with expression for x_n (once I realised that it should be n! not n in the denominator) is that if you expand out the n! you get


    This is just a geometric mean. Maybe that helps you?
    Last edited: Feb 9, 2008
  5. Feb 9, 2008 #4


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    might be easier to take logs, so
    \log x_n = \sum_{k=1}^{n} (-\log(k/n)) (1/n)
    which will approach the limit [itex]\int_0^1(-\log(x))dx[/itex] from below. You still need to show that the approximations converge monotonically to the limit as n increases.
  6. Feb 9, 2008 #5
    Thank you, gel, I have corrected x_n with ! in my first post. Also, I have found error in my original post:
    [tex]\lim_{n\rightarrow\infty}z_n=e \Longrightarrow \lim_{n\rightarrow\infty}x_n=e[/tex]

    Should be: [tex]\lim_{n\rightarrow\infty}z_n=\frac{1}{e} \Longrightarrow e\geq \lim_{n\rightarrow\infty}x_n\geq \frac{1}{e}[/tex]

    So, now I have proven [tex]x_n[/tex] is bounded, but not the convergence to e :-(
    I'll think about geometric mean and logarithms, maybe I will come up with something :-)
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