Question Number 215737 by Ismoiljon_008 last updated on 16/Jan/25
Commented by Ismoiljon_008 last updated on 16/Jan/25
$$\:\:\:{compare}\:{these}\:{two}\:{numbers} \\ $$$$\:\:\:{help}\:{me},\:{please} \\ $$
Answered by A5T last updated on 16/Jan/25
$$\frac{\left(\mathrm{n}!\right)^{\frac{\mathrm{1}}{\mathrm{n}}} }{\left(\mathrm{n}+\mathrm{1}\right)!^{\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}} }=\mathrm{x}\:\left(\mathrm{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{integer}\right) \\ $$$$\mathrm{x}^{\mathrm{n}} =\frac{\mathrm{n}!}{\left(\mathrm{n}+\mathrm{1}\right)!^{\frac{\mathrm{n}}{\mathrm{n}+\mathrm{1}}} };\:\mathrm{x}^{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)} =\frac{\left(\mathrm{n}!\right)^{\mathrm{n}+\mathrm{1}} }{\left(\left(\mathrm{n}+\mathrm{1}\right)!\right)^{\mathrm{n}} }=\frac{\left(\mathrm{n}!\right)^{\mathrm{n}} ×\mathrm{n}!}{=\left(\mathrm{n}!\right)^{\mathrm{n}} \left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}} } \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)} =\frac{\mathrm{n}!}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}} }<\mathrm{1}\:\mathrm{where}\:\mathrm{n}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{integer} \\ $$$$\Rightarrow\mathrm{x}<\mathrm{1}\Rightarrow\left(\mathrm{n}!\right)^{\frac{\mathrm{1}}{\mathrm{n}}} <\left(\mathrm{n}+\mathrm{1}\right)!^{\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}} \\ $$$$\Rightarrow\sqrt[{\mathrm{2006}}]{\mathrm{2006}!}<\sqrt[{\mathrm{2007}}]{\mathrm{2007}!} \\ $$
Commented by Ismoiljon_008 last updated on 17/Jan/25
$$\:\:\:{thank}\:{you}\:{very}\:{much} \\ $$
Answered by mr W last updated on 16/Jan/25
![for n≥2: n!=1×2×3×...×n <n×n×n×...×n=n^n (n!)^(1/n) <(n^n )^(1/n) =n say f(n)=(n!)^(1/n) ((f(n+1))/(f(n)))=(([(n+1)!]^(1/(n+1)) )/((n!)^(1/n) )) =(([(n+1)n!]^(1/(n+1)) )/([(n!)^((n+1)/n) ]^(1/(n+1)) ))=[((n+1)/((n!)^(1/n) ))]^(1/(n+1)) >[((n+1)/((n^n )^(1/n) ))]^(1/(n+1)) =(((n+1)/n))^(1/(n+1)) =(1+(1/n))^(1/(n+1)) >1+(1/(n(n+1)))>1 ⇒f(n+1)>f(n) ⇒f(n)<f(n+1) i.e. ((n!))^(1/n) <(((n+1)!))^(1/(n+1)) therefore ((2006!))^(1/(2006)) <((2007!))^(1/(2007))](https://www.tinkutara.com/question/Q215742.png)
$${for}\:{n}\geqslant\mathrm{2}: \\ $$$${n}!=\mathrm{1}×\mathrm{2}×\mathrm{3}×…×{n} \\ $$$$\:\:\:\:\:<{n}×{n}×{n}×…×{n}={n}^{{n}} \\ $$$$\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} <\left({n}^{{n}} \right)^{\frac{\mathrm{1}}{{n}}} ={n} \\ $$$$ \\ $$$${say}\:{f}\left({n}\right)=\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} \\ $$$$\frac{{f}\left({n}+\mathrm{1}\right)}{{f}\left({n}\right)}=\frac{\left[\left({n}+\mathrm{1}\right)!\right]^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} }{\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} } \\ $$$$\:\:=\frac{\left[\left({n}+\mathrm{1}\right){n}!\right]^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} }{\left[\left({n}!\right)^{\frac{{n}+\mathrm{1}}{{n}}} \right]^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} }=\left[\frac{{n}+\mathrm{1}}{\left({n}!\right)^{\frac{\mathrm{1}}{{n}}} }\right]^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} \\ $$$$\:\:>\left[\frac{{n}+\mathrm{1}}{\left({n}^{{n}} \right)^{\frac{\mathrm{1}}{{n}}} }\right]^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} =\left(\frac{{n}+\mathrm{1}}{{n}}\right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} =\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)^{\frac{\mathrm{1}}{{n}+\mathrm{1}}} \\ $$$$\:>\mathrm{1}+\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}>\mathrm{1} \\ $$$$\Rightarrow{f}\left({n}+\mathrm{1}\right)>{f}\left({n}\right)\: \\ $$$$\Rightarrow{f}\left({n}\right)<{f}\left({n}+\mathrm{1}\right) \\ $$$${i}.{e}.\:\sqrt[{{n}}]{{n}!}<\sqrt[{{n}+\mathrm{1}}]{\left({n}+\mathrm{1}\right)!} \\ $$$${therefore}\:\sqrt[{\mathrm{2006}}]{\mathrm{2006}!}<\sqrt[{\mathrm{2007}}]{\mathrm{2007}!} \\ $$
Commented by Ismoiljon_008 last updated on 17/Jan/25
$$\:\:\:{thank}\:{you}\:{very}\:{much} \\ $$