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Math Help - Proof

  1. #1
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    Proof

    if f(1) + f(2) +... + f(n-1)< the integral from 1 to n of f(x)dx < f(2) + f(3) +...+ f(n).
    Chose f(x) = ln x. Show that n^n/e^(n-1)< n! < n^(n+1)/e^n
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Snooks02 View Post
    if f(1) + f(2) +... + f(n-1)< the integral from 1 to n of f(x)dx < f(2) + f(3) +...+ f(n).
    Chose f(x) = ln x. Show that n^n/e^(n-1)< n! < n^(n+1)/e^n
    Assume that (n!)'>0 and n>0

    \begin{aligned}\sum_{k=1}^{n-1}\ln(k)&=\ln(1)+\ln(2)+\cdots+\ln(n-1)\\<br />
&=\ln\left(1\cdot{2}\cdots{n-1}\right)\\<br />
&=\ln((n-1)!)\end{aligned}

    And

    \int_1^n\ln(x)dx=n\ln(n)-n+1=\ln\left(\frac{n^n}{e^{n-1}}\right)

    As well as

    \begin{aligned}\sum_{k=2}^{n}\ln(k)&=\ln(2)+\ln(3)  +\cdots+\ln(n)\\<br />
&=\ln\left(1\cdot{2}\cdots{n}\right)\\<br />
&=\ln(n!)\end{aligned}

    So we would have that

    \ln((n-1)!)<\ln\left(\frac{n^n}{e^{n-1}}\right)<\ln(n!)

    And since the exponential function is stricly increasing across its domain the above inequality also gives

    (n-1)!<\frac{n^n}{e^{n-1}}<n!

    So then because n! is strictly increasing we may rewrite our inequality as

    1<\frac{n^n}{e^{n-1}}\cdot\frac{1}{(n-1)!}<n

    Reciporcating and not forgetting to reverse the direction of the inequalities gives

    \frac{1}{n}<(n-1)!\cdot\frac{e^{n-1}}{n^n}<1

    Multiplying through by \frac{n^n}{e^{n-1}} gives

    \frac{n^{n-1}}{e^{n-1}}<(n-1)!<\frac{n^n}{e^{n-1}}

    Finally multiplying through by n gives

    \frac{n^n}{e^{n-1}}<n!<\frac{n^{n+1}}{e^{n-1}}\quad\blacksquare
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