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

$$prove$$ $${\left(\frac{n+1}{3}\right)}^n<n!$$

Commented bymr W last updated on 19/Jul/22

$$goodideasir!thanks!$$ $$n!\approx \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n$$ $$\underset{n\to \mathrm{\infty }}{\lim }\frac{n+1}{\sqrt[n]{n!}}$$ $$=\underset{n\to \mathrm{\infty }}{\lim }\frac{(n+1)}{{\left(2n\pi \right)}^{\frac{1}{2n}}\left(\frac{n}{e}\right)}$$ $$=\underset{n\to \mathrm{\infty }}{\lim }\frac{e}{{\left(\frac{\pi }{\frac{1}{2n}}\right)}^{\frac{1}{2n}}}(1+\frac{1}{n})$$ $$=\underset{n\to \mathrm{\infty }}{\lim }\frac{e}{{\left(\frac{\pi }{\frac{1}{2n}}\right)}^{\frac{1}{2n}}}$$ $$=\underset{x\to 0}{\lim }\frac{e}{{\left(\frac{\pi }{x}\right)}^x}$$ $$=\frac{e}{1}$$ $$=e<3$$

Commented byFrix last updated on 18/Jul/22

$$\iff$$ $$3>\frac{n+1}{\sqrt[n]{n!}}$$ $$\underset{n\to \mathrm{\infty }}{\lim }\frac{n+1}{\sqrt[n]{n!}}=\mathrm{e}<3$$ $$\mathrm{just}\mathrm{a}\mathrm{idea}...$$

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