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December 23rd, 2018, 12:26 PM   #1
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Range of x

Find range of $\displaystyle x$ such that $\displaystyle x^n n! \geq n^n$
$\displaystyle n\in N$ and $\displaystyle x$ -real number
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December 24th, 2018, 06:00 AM   #2
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What about integral inequality ? Using euler integral
$\displaystyle \int_{0}^{\infty}x^n e^{-x}dx> \int_{n}^{\infty} x^n e^{-x}dx >n^n \int_{n}^{\infty} e^{-x} dx$
Maybe there are other ways but i tried integrals

Last edited by idontknow; December 24th, 2018 at 06:18 AM.
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December 24th, 2018, 08:23 AM   #3
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Did you try applying the bounds on n! given by Stirling's formula?
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December 24th, 2018, 12:30 PM   #4
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The range of x will be dependent on n. For example $n=1,\ x\ge 1$, for $n=2,\ x\ge \sqrt{2}$.

General formula $x\ge \frac{n}{(n!)^{1/n}}$ Use Stirling's formula to estimate for large n.
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Last edited by mathman; December 24th, 2018 at 12:40 PM.
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December 25th, 2018, 03:45 AM   #5
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I got x>e , am i correct ?
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December 25th, 2018, 01:16 PM   #6
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Quote:
Originally Posted by idontknow View Post
I got x>e , am i correct ?
As long as you want one bound for all n, this is correct.
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