Prove that n!>((n/3))^n


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Question Number 13728 by prakash jain last updated on 22/May/17

$Prove that$ $n! > {(\frac{n}{3})}^{n}$
	Answered by mrW1 last updated on 23/May/17

	$U s i n g m a t h e m a t i c a l i n d u c t i o n :$ $f o r n = 1 w e h a v e$ $1! = 1 > {(\frac{1}{3})}^{1} = \frac{1}{3} \Rightarrow t r u e$ $s u p p o s e d {i t}^{'} s t r u e f o r n, i . e .$ $n! > {(\frac{n}{3})}^{n}$ $f o r n + 1 w e h a v e$ $(n + 1)! = (n + 1) n! > (n + 1) {(\frac{n}{3})}^{n}$ $= {(\frac{n + 1}{3})}^{n + 1} (\frac{3}{n + 1}) {(\frac{3}{n + 1} \times \frac{n}{3})}^{n} (n + 1)$ $= {(\frac{n + 1}{3})}^{n + 1} 3 {(\frac{n}{n + 1})}^{n} > {(\frac{n + 1}{3})}^{n + 1}$ $s i n c e 3 {(\frac{n}{n + 1})}^{n} > 1 (* s e e p r o o f)$ $s o {i t}^{'} s a l s o t r u e f o r n + 1 .$ $\Rightarrow {i t}^{'} s t r u e f o r a l l n .$
	Commented by prakash jain last updated on 23/May/17

	$This is great .$ $Proof for$ $3 {(\frac{n}{n + 1})}^{n} > 1$ $We can see that {(\frac{n}{n + 1})}^{n} > {(\frac{n + 1}{n + 2})}^{n + 1} and$ $\lim_{n \to \infty} {(\frac{n}{1 + n})}^{n} = \frac{1}{e}$ $\frac{3}{e} > 1$
	Commented by mrW1 last updated on 23/May/17

	${T h a t}^{'} s g o o d i d e a!$ $I w a n t e d t o t r y t o p r o v e 3 {(\frac{n}{n + 1})}^{n} > 1 a l s o u s i n g$ $m a t h e m a t i c a l i n d u c t i o n .$
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