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prove that $$(\frac{n}{3})^n<n!<e\cdot(\frac{n}{2})^n$$

I tried to prove by the induction that $(\frac{n}{3})^n<n!$ and $n!<e\cdot(\frac{n}{2})^n$, but I failed

my assumption $n^n<n!*3^n$

$$(n+1)^{n+1}<(n+1)*3^{n+1}$$ $$\frac{(n+1)^n}{9}<n!*3^{n-1}$$

edited May 01 '16 at 15:37

Martin Sleziak

asked Nov 22 '13 at 09:50

Kran

You failed where exactly? How far you got? – ulead86 Nov 22 '13 at 09:54
You probably could use Stirling approximation for Log[n!] – Claude Leibovici Nov 22 '13 at 10:10
and i don't understand why it doesn't work fo n=1 – Kran Nov 22 '13 at 10:17
@Kacper for n = 1 you have $\frac{1}{3} < 1 < \frac{e}{2}$ which is true – oks Nov 22 '13 at 10:47

prove that $(\frac{n}{3})^n

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