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Thread: Today's calculation of integral #8

  1. #1
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    Today's calculation of integral #8

    Compute $\displaystyle \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}.$
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Compute $\displaystyle \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}.$
    let $\displaystyle a_n = \frac {n^n}{n!}$ and note that the root test will give the same result as the ratio test. that is, we will use the fact that:

    $\displaystyle \lim |a_n|^{1/n} = \lim \left| \frac {a_{n + 1}}{a_n} \right|$

    Now compute $\displaystyle \lim \left| \frac {a_{n + 1}}{a_n} \right|$

    if i did not make a mistake, the answer should be $\displaystyle e$
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    Yeah, that's the answer, but it is asked a solution which involves integration
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Yeah, that's the answer, but it is asked a solution which involves integration
    well, you're the Inte-Killer, not me.

    I have to go for a while. I'll think about it and maye post a solution with integrals later... maybe
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    Here is a way to do it. But it might be cheating, because it is basically Stirling approximation.

    Instead compute the $\displaystyle \ln $ of this sequence,
    $\displaystyle \ln \left( \frac{n^n}{n!} \right)^{1/n} = \ln n - \frac{1}{n}\ln (n!)$.

    But, $\displaystyle \ln (n!) = \sum_{k=1}^n \ln k \approx \int_1^n \ln x dx = n\ln n - n + 1$ by Euler-Maclaurin formula, the remainder term will happen to go to zero.

    Thus, we are left with $\displaystyle 1 - \frac{1}{n} \to 1$. Which means $\displaystyle \ln a_n \to 1 \implies a_n \to e$.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Compute $\displaystyle \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}.$
    $\displaystyle y=\lim_{n \to {\infty}}\bigg(\frac{n^{n}}{n!}\bigg)^{\frac{1}{n} }$....so $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{1}{n}\cdot\ln\bigg(\frac{n^n}{n!}\b igg)$...$\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{1}{n}ln\bigg(\cdot\frac{n^{n}}{\int _0^{\infty}e^{-x}x^{n}dx}\bigg)$...using l'hopital's rule and much simplification we get $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{ln(n)n!+n!-\int_0^{\infty}e^{-x}x^n\ln(x)dx}{n!}$...using l'hopitals again gives us$\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{ln(n)e^{-n}n^n+(n-1)!+e^{-n}n^n-ln(n)e^{-n}n^n}{e^{-n}n^{n}}$..simplifying we get $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!+e^{-n}n^n}{e^{-n}n^n}$...now by splitting we get $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!}{e^{-n}n^n}+1=0+1$...so $\displaystyle ln(y)=1$...so $\displaystyle y=e$
    Last edited by Mathstud28; Apr 10th 2008 at 07:02 PM.
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    Okay, let this be more interesting: no powerful theorems.

    As for Mathstud28's solution, no L'H˘pital's Rule of course.
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    Super Member angel.white's Avatar
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    How do we know that we can treat this as an integral? Like how did you know to use the ratio test? I was looking at it as just a limit that (I'm pretty sure) goes to infinity.

    :/ I don't think I even understand the question.


    Edit: or... goes to e :P
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    Haha

    Quote Originally Posted by Krizalid View Post
    Okay, let this be more interesting: no powerful theorems.

    As for Mathstud28's solution, no L'H˘pital's Rule of course.
    Oh...come on I am only in eleventh grade can't I get soem credit haha...that took me forever haha....thanks for nothing ....haha I am just kidding that was hard for me though....I am happy with myself
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    Just wondering

    Quote Originally Posted by Mathstud28 View Post
    $\displaystyle y=\lim_{n \to {\infty}}\bigg(\frac{n^{n}}{n!}\bigg)^{\frac{1}{n} }$....so $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{1}{n}\cdot\ln\bigg(\frac{n^n}{n!}\b igg)$...$\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{1}{n}ln\bigg(\cdot\frac{n^{n}}{\int _0^{\infty}e^{-x}x^{n}dx}\bigg)$...using l'hopital's rule and much simplification we get $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{ln(n)n!+n!-\int_0^{\infty}e^{-x}x^n\ln(x)dx}{n!}$...using l'hopitals again gives us$\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{ln(n)e^{-n}n^n+(n-1)!+e^{-n}n^n-ln(n)e^{-n}n^n}{e^{-n}n^{n}}$..simplifying we get $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!+e^{-n}n^n}{e^{-n}n^n}$...now by splitting we get $\displaystyle ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!}{e^{-n}n^n}+1=0+1$...so $\displaystyle ln(y)=1$...so $\displaystyle y=e$
    But is this right though?
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  11. #11
    Moo
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    Where Krizalid loves integrals, mathstud28 is fond of l'H˘pital's rule

    I would have done like TPH, but i don't see where there is a "powerful theorem" ?
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    Quote Originally Posted by Moo View Post
    I would have done like TPH, but i don't see where there is a "powerful theorem" ?
    The powerful theorem here is the Euler-Maclurin summation which tells us that we can approximate a sum with an integral with error estimate good enough to vanish in the limit.
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    It appears to me all you need to do is put a bound on $\displaystyle n!$. Of course, Stirling approximation is the best way to go, but that would be cheating here. You can use what PaulRS did here to get good estimates on $\displaystyle n!$ and use squeeze theorem.
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    Quote Originally Posted by Krizalid View Post
    Compute $\displaystyle \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}.$
    $\displaystyle \left\{ {\frac{{n^n }}
    {{n!}}} \right\}^{\frac{1}
    {n}} = \exp \left\{ {\frac{1}
    {n}\ln \frac{{n^n }}
    {{n!}}} \right\} = \exp \left\{ {\frac{1}
    {n}\ln \left( {\prod\limits_{k\, = \,1}^n {\frac{n}
    {k}} } \right)} \right\} = \exp \left\{ { - \frac{1}
    {n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}
    {n}} } \right\}.$

    As $\displaystyle n\to\infty,\,{\frac{1}
    {n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}
    {n}} }$ is a Riemann sum for $\displaystyle \int_0^1\ln x\,dx,$ which is a fairly known integral, it follows that $\displaystyle \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}=e.$
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle \left\{ {\frac{{n^n }}
    {{n!}}} \right\}^{\frac{1}
    {n}} = \exp \left\{ {\frac{1}
    {n}\ln \frac{{n^n }}
    {{n!}}} \right\} = \exp \left\{ {\frac{1}
    {n}\ln \left( {\prod\limits_{k\, = \,1}^n {\frac{n}
    {k}} } \right)} \right\} = \exp \left\{ { - \frac{1}
    {n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}
    {n}} } \right\}.$

    As $\displaystyle n\to\infty,\,{\frac{1}
    {n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}
    {n}} }$ is a Riemann sum for $\displaystyle \int_0^1\ln x\,dx,$ which is a fairly known integral, it follows that $\displaystyle \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac 1n}=e.$
    Cool solution. Is that your own problem?
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