prove-n-1-3-n-lt-n-

Question Number 173815 by mr W last updated on 18/Jul/22

p r o v e

{(\frac{n + 1}{3})}^{n} < n!

Commented by mr W last updated on 19/Jul/22

g o o d i d e a s i r! t h a n k s!

n! \approx \sqrt{2 π n} {(\frac{n}{e})}^{n}

\lim_{n \to \infty} \frac{n + 1}{\sqrt[n]{n!}}

= \lim_{n \to \infty} \frac{(n + 1)}{{(2 n π)}^{\frac{1}{2 n}} (\frac{n}{e})}

= \lim_{n \to \infty} \frac{e}{{(\frac{π}{\frac{1}{2 n}})}^{\frac{1}{2 n}}} (1 + \frac{1}{n})

= \lim_{n \to \infty} \frac{e}{{(\frac{π}{\frac{1}{2 n}})}^{\frac{1}{2 n}}}

= \lim_{x \to 0} \frac{e}{{(\frac{π}{x})}^{x}}

= \frac{e}{1}

= e < 3

Commented by Frix last updated on 18/Jul/22

\Leftrightarrow

3 > \frac{n + 1}{\sqrt[n]{n!}}

\lim_{n \to \infty} \frac{n + 1}{\sqrt[n]{n!}} = e < 3

just a idea \dots

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