December 23rd, 2018, 12:26 PM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68  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: iPhone Posts: 436 Thanks: 68 
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: 2,371 Thanks: 1275 
Did you try applying the bounds on n! given by Stirling's formula?

December 24th, 2018, 12:30 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
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. Last edited by mathman; December 24th, 2018 at 12:40 PM. 
December 25th, 2018, 03:45 AM  #5 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
I got x>e , am i correct ?

December 25th, 2018, 01:16 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670  

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