Prove-that-n-gt-n-3-n-

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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