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 December 23rd, 2018, 12:26 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 606 Thanks: 88 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
 December 24th, 2018, 06:00 AM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 606 Thanks: 88 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.
 December 24th, 2018, 08:23 AM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,533 Thanks: 1390 Did you try applying the bounds on n! given by Stirling's formula? Thanks from topsquark
 December 24th, 2018, 12:30 PM #4 Global Moderator   Joined: May 2007 Posts: 6,807 Thanks: 717 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. Thanks from topsquark and idontknow Last edited by mathman; December 24th, 2018 at 12:40 PM.
 December 25th, 2018, 03:45 AM #5 Senior Member   Joined: Dec 2015 From: somewhere Posts: 606 Thanks: 88 I got x>e , am i correct ?
December 25th, 2018, 01:16 PM   #6
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Quote:
 Originally Posted by idontknow I got x>e , am i correct ?
As long as you want one bound for all n, this is correct.

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