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This is what I am trying to solve:

$n!>n^{\frac {n}{2}}$

I tried with induction and somehow it doesn`t work this way, I tried with logarithms and also didn´t find a way. Is there an elementary(if possible) way to prove this?

Or is it the case that for some large $n_0$ this inequality reverses sign?

Martin Sleziak
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    You may want to look at the Wikipedia article on Stirling's Formula. That will answer all possible questions. But the question you asked below about replacing $2$ by $x$ can be answered more simply, by small modification of the proof for $x=2$. – André Nicolas Mar 23 '13 at 20:19
  • I believed that it works for every $1<x<2$ but it looks very hard for me to find an elementary answer, because I am also some kind of a beginner in mathematics. Could it be that there exist some combinatorial explanation of the fact that it is true? –  Mar 23 '13 at 20:27
  • Thus, if you want Andre to be notified of your comment you should begin with @AndréNicolas . Naturally, there is no real time limit on the next time he looks at his notifications, so it could still take time. – Will Jagy Mar 23 '13 at 20:31
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    You can modify Will Jagy's argument. Or you can do a weak Sirling formula by estimating $\log 2+\log 3+\cdots +\log n$ using an integral. I do not know any argument that is purely combinatorial. – André Nicolas Mar 23 '13 at 20:33
  • Do you want to say that if I don`t use @André Nicolas or @Will Jagy at the beginning of a comment that then the notifications are not coming to you? –  Mar 23 '13 at 20:35
  • @Will Jagy If that is true I didn´t know till now! –  Mar 23 '13 at 20:41
  • Thus, yes, I was notified of the one with the at sign and my name,but not the one befroe (there is some problem with more than one at sign). If I click on my name to see my profile, the word "Responses" will show a number, one or more, and if I click on responses it will display them. If I do not click on my profile for an hour or a day I remain unaware. You should be notified of this because it is directly under a question by you. Directly under an answer by you would also work. – Will Jagy Mar 23 '13 at 20:48
  • See if you can do Andre's variation with exponent $2n/3.$ – Will Jagy Mar 23 '13 at 20:50
  • Oh, even without Stirling, the same technique shows rapidly that $n^n > n!$ for $n > 1.$ So if the base and the exponent were, say, $7 n / 8,$ I would expect that to be larger for small $n,$ then $n!$ getting bigger for bigger $n.$ – Will Jagy Mar 23 '13 at 23:00
  • @Will Jagy I imagined that when it is shown that above mentioned inequality is true for exponents that are rational multiples $qn$ such that $q<1$ and $q$ is rational then there must be some principle to expand it to all real multiples $xn$, $x<1$. –  Mar 23 '13 at 23:15
  • Sounds correct. If $q < 1$ is rational, with real $x$ and $0 < q < x < 1,$ then $$ (qn)^{qn} < (xn)^{xn} $$ for all natural numbers $n,$ inequality reversed if $0 < x < q < 1.$ – Will Jagy Mar 23 '13 at 23:38
  • @Will Jagy Also, if you could teach me how to put the link of something to the question so that it doesn`t appear as the link but has the appereance in blue letters and instead of the link there stand words describing what the link is about, I see that everywhere on this site but do not know how to produce it? –  Mar 26 '13 at 13:55
  • ...which is the sine function –  Mar 26 '13 at 15:34
  • I found it, sorry for disturbing you! –  Mar 26 '13 at 15:35
  • This is basically the same question as: http://math.stackexchange.com/questions/785514/solving-an-inequality-n-geq-3-nn-lt-n2 – Martin Sleziak Jan 21 '15 at 17:18

2 Answers2

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For $n$ even, you get $n! = (n/2)! \cdot B,$ where $B$ is the product of the $(n/2)$ numbers from $1 +(n/2)$ to $n.$ So, $$ B > (n/2)^{(n/2)} = \frac{n^{n/2}}{2^{n/2}}. $$ Then you need $$(n/2)! > 2^{n/2} $$ which is not difficult by induction.

Mild revisions for odd $n$

Will Jagy
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  • This is elegant, I am wondering now could we change $2$ in $n!>n^{\frac {n}{2}}$ by any $1<x<2$ so that for sufficiently large $n$ the equality holds for that $x$? I would appreciate if you would explain to me why or why not in the comment or in the answer so that I do not pose another question. I strongly accept this as an answer. –  Mar 23 '13 at 20:05
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    Yes, for $\delta > 0$ and sufficiently large $n,$ we do get $$ n! > n^{\frac{n}{1 + \delta}}, $$ by http://en.wikipedia.org/wiki/Stirling%27s_approximation It is hard to imagine an elementary approach, though. – Will Jagy Mar 23 '13 at 20:19
  • Apparently Andre knows how to replace 2 by in this by simple means, I do not see it yet. – Will Jagy Mar 23 '13 at 20:21
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$(n!)^2=\{1.2.3......n\}\{n.(n-1)......3.2.1\}=\{(1.n)(2.(n-1))........(n.1)\} \geq n.n.n.....n (\text{n}\space times) =n^n \implies n! \geq (n)^{n/2}$.

learner
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