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I want to prove that $x_n=\frac{\sqrt[n]{n!}}{n}$ is decreasing, but I tried everything I know and it didn't work. At first, I tried to prove $x_n\ge x_{n+1}$ directly which is equivalent with $(n+1)^{n^2}n!\ge n^{n(n+1)}$. Then, I tried to prove that using induction, but it didn't work. I tried to use Stirling Formula too, in order to approximate $n!$ but that led me to $(1+1/n)^n\ge e$ which is clearly false. Any thoughts ? Please, I really need to solve it.

Gary
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Lucas McAllister
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  • TRy Stirling approximation. – Claude Leibovici Feb 13 '22 at 08:51
  • I tried and it didn't work. I've already told that it led me to $(1+1/n)^n \geqslant e$, which is false. – Lucas McAllister Feb 13 '22 at 08:54
  • We have $$ x_n > x_{n + 1} \Longleftrightarrow \left( {1 + \frac{1}{n}} \right)^{n(n + 1)} n! >n^n \left( {1 + \frac{1}{n}} \right)^n . $$ But $$ \left( {1 + \frac{1}{n}} \right)^{n(n + 1)} n! > e^n n!,\quad n^n e > n^n \left( {1 + \frac{1}{n}} \right)^n . $$ Thus it is enough to show $$ n! \ge \left( {\frac{n}{e}} \right)^n e. $$ which is true for $n\geq 1$. – Gary Feb 13 '22 at 08:55
  • It works like a charm – Claude Leibovici Feb 13 '22 at 08:56
  • @ClaudeLeibovici Demonstrate it by posting a detaild answer! :) – Gary Feb 13 '22 at 08:57
  • @Gary, well, using Stirling's approximation, your last inequality seems true. I hope I'm not wrong. Thanks ! – Lucas McAllister Feb 13 '22 at 09:11
  • Stirling would comfirm it for large $n$ without error bounds. It is known however that $n!>(n/e)^n \sqrt{2\pi n}$ for all $n\geq 1$, i.e., Stirling's approximation is always a lower bound. It is possible to prove the inequality $n! \ge (n/e)^n e$ without this. – Gary Feb 13 '22 at 09:13
  • @Gary But by using Stirling's approximation is enough to prove that $2πn>=e^2$ which is true for $n>=2$ and for me is enough. I don't say it is not possible in other way, but I have no idea how. – Lucas McAllister Feb 13 '22 at 09:31
  • @Gary Ok, you were right, I managed to prove it by using induction and without Stirling's approximation. – Lucas McAllister Feb 13 '22 at 09:40
  • @LucasMcAllister Induction is better because Stirling's formula is an approximation. So you would need $$ n! \ge \left( {\frac{n}{e}} \right)^n \sqrt {2\pi n} $$ otherwise it may not be true that $2\pi n\geq e^2$ is sufficient. I hope you understand my point. Stirling would gurantee that the inequality you need to show is true for sufficiently large values of $n$. But you need it for all $n\geq 1$. It is true that Stirling's formula is also a lower bound but it is not a necessary consequence of the approximation. – Gary Feb 13 '22 at 10:02
  • @Gary Yeah, I got your point. Thanks for your help! – Lucas McAllister Feb 13 '22 at 11:19

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