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Math Help - Today's calculation of integral #8

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    Today's calculation of integral #8

    Compute \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 \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac  1n}.
    let 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:

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

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

    if i did not make a mistake, the answer should be 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 \ln of this sequence,
    \ln \left( \frac{n^n}{n!} \right)^{1/n} = \ln n - \frac{1}{n}\ln (n!).

    But, \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 1 - \frac{1}{n} \to 1. Which means  \ln a_n \to 1 \implies a_n \to e.
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    Quote Originally Posted by Krizalid View Post
    Compute \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac  1n}.
    y=\lim_{n \to {\infty}}\bigg(\frac{n^{n}}{n!}\bigg)^{\frac{1}{n}  }....so ln(y)=\lim_{n \to {\infty}}\frac{1}{n}\cdot\ln\bigg(\frac{n^n}{n!}\b  igg)... 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 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 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 ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!+e^{-n}n^n}{e^{-n}n^n}...now by splitting we get ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!}{e^{-n}n^n}+1=0+1...so ln(y)=1...so y=e
    Last edited by Mathstud28; April 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
    y=\lim_{n \to {\infty}}\bigg(\frac{n^{n}}{n!}\bigg)^{\frac{1}{n}  }....so ln(y)=\lim_{n \to {\infty}}\frac{1}{n}\cdot\ln\bigg(\frac{n^n}{n!}\b  igg)... 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 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 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 ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!+e^{-n}n^n}{e^{-n}n^n}...now by splitting we get ln(y)=\lim_{n \to {\infty}}\frac{(n-1)!}{e^{-n}n^n}+1=0+1...so ln(y)=1...so 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 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 n! and use squeeze theorem.
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    Quote Originally Posted by Krizalid View Post
    Compute \lim_{n\to\infty}\bigg(\frac{n^n}{n!}\bigg)^{\frac  1n}.
    \left\{ {\frac{{n^n }}<br />
{{n!}}} \right\}^{\frac{1}<br />
{n}}  = \exp \left\{ {\frac{1}<br />
{n}\ln \frac{{n^n }}<br />
{{n!}}} \right\} = \exp \left\{ {\frac{1}<br />
{n}\ln \left( {\prod\limits_{k\, = \,1}^n {\frac{n}<br />
{k}} } \right)} \right\} = \exp \left\{ { - \frac{1}<br />
{n}\sum\limits_{k\, = \,1}^n {\ln \frac{k}<br />
{n}} } \right\}.

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

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