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Question Number 172560 by mr W last updated on 28/Jun/22

prove for n∈N and n>1  (((n+1)/3))^n <n!<(((n+1)/2))^n

$${prove}\:{for}\:{n}\in{N}\:{and}\:{n}>\mathrm{1} \\ $$ $$\left(\frac{{n}+\mathrm{1}}{\mathrm{3}}\right)^{{n}} <{n}!<\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\right)^{{n}} \\ $$

Commented bymr W last updated on 07/Sep/22

see also Q173815

$${see}\:{also}\:{Q}\mathrm{173815} \\ $$

Answered by Jamshidbek last updated on 28/Jun/22

Hint: Mathematic induction method

$$\mathrm{Hint}:\:\mathrm{Mathematic}\:\mathrm{induction}\:\mathrm{method} \\ $$

Answered by mnjuly1970 last updated on 28/Jun/22

   (n>1) :(( 1+2+...+n)/n) >((1.2.3...n))^(1/n)          ((n+1)/2) >((n!))^(1/n)  ⇒ (((n+1)/2) )^( n) >n!

$$\:\:\:\left({n}>\mathrm{1}\right)\::\frac{\:\mathrm{1}+\mathrm{2}+...+{n}}{{n}}\:>\sqrt[{{n}}]{\mathrm{1}.\mathrm{2}.\mathrm{3}...{n}} \\ $$ $$\:\:\:\:\:\:\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\:>\sqrt[{{n}}]{{n}!}\:\Rightarrow\:\left(\frac{{n}+\mathrm{1}}{\mathrm{2}}\:\right)^{\:{n}} >{n}! \\ $$

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