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

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 $n^n<n!*3^n$

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

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You failed where exactly? How far you got? –  ulead86 Nov 22 '13 at 9:54
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 –  Kacper 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
$$(n+1)^{n+1} \\ = \left(\frac{n+1}{n}\right)^n\ n^{n}\ (n+1) \\ = \left( 1+\frac{1}{n} \right)^n\ n^{n}\ (n+1) \\ < e \ n^{n} \ (n+1) \mbox{ because \left( 1+\frac{1}{n} \right)^n increases and tends to e}\\ < e \ 3^{n} n! \ (n+1) \mbox{ by inductive hypothesis about n^n }\\ < 3 \times \ 3^{n} (n+1)! \\ = 3^{n+1} (n+1)!$$