Question Number 79932 by mr W last updated on 29/Jan/20
Commented by mr W last updated on 29/Jan/20
$$\left[{Q}\mathrm{79861}\:{reposted}\right] \\ $$
Answered by mr W last updated on 29/Jan/20
$${I}\:{think}\:{I}'{ve}\:{found}\:{an}\:{easy}\:{way}\:{to}\:{prove}: \\ $$$$ \\ $$$${let}'{s}\:{look}\:{at}\:{f}\left({x}\right)=\mathrm{ln}\:\left({x}\right),\:{we}\:{see} \\ $$$${f}\left({x}\right)=\mathrm{ln}\:{x}\leqslant\frac{{x}}{{e}}\:\left(“=''\:{is}\:{only}\:{at}\:{x}={e}\right) \\ $$$$\left({see}\:{diagram}\:{below}\:{if}\:{you}\:{have}\:{doubt}\right) \\ $$$$ \\ $$$${P}_{{n}} =\sqrt{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{4}}]{\mathrm{4}…\sqrt[{{n}}]{{n}}}}} \\ $$$${P}_{{n}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{4}^{\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{4}}} …{n}^{\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{4}}×…×\frac{\mathrm{1}}{{n}}} \\ $$$${P}_{{n}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}!}} \mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}!}} \mathrm{4}^{\frac{\mathrm{1}}{\mathrm{4}!}} …{n}^{\frac{\mathrm{1}}{{n}!}} =\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}{k}^{\frac{\mathrm{1}}{{k}!}} \\ $$$$\mathrm{ln}\:{P}_{{n}} =\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\mathrm{ln}\:{k}}{{k}!}<\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\:\frac{{k}}{{ek}!}=\frac{\mathrm{1}}{{e}}\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\left({k}−\mathrm{1}\right)!}=\frac{\mathrm{1}}{{e}}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}!} \\ $$$$<\frac{\mathrm{1}}{{e}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}!}=\frac{\mathrm{1}}{{e}}\left(\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}!}−\mathrm{1}\right)=\frac{\mathrm{1}}{{e}}\left({e}−\mathrm{1}\right)=\mathrm{1}−\frac{\mathrm{1}}{{e}} \\ $$$$\Rightarrow{P}_{{n}} <{e}^{\mathrm{1}−\frac{\mathrm{1}}{{e}}} =\frac{{e}}{\:\sqrt[{{e}}]{{e}}}\approx\mathrm{1}.\mathrm{8816}<\mathrm{2}\:\Rightarrow{proved} \\ $$
Commented by M±th+et£s last updated on 29/Jan/20
$${thank}\:{you}\:{sir}\:{great}\:{solution} \\ $$
Commented by MJS last updated on 29/Jan/20
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{P}_{{n}} \:\approx\mathrm{1}.\mathrm{82902467956} \\ $$
Commented by MJS last updated on 29/Jan/20
$${P}_{{n}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}!}} \mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}!}} \mathrm{4}^{\frac{\mathrm{1}}{\mathrm{4}!}} …{n}^{\frac{\mathrm{1}}{{n}!}} \\ $$$$\mathrm{looking}\:\mathrm{at}\:\mathrm{the}\:\mathrm{prime}\:\mathrm{factors}… \\ $$$$\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{2}!}} \mathrm{3}^{\frac{\mathrm{1}}{\mathrm{3}!}} \mathrm{4}^{\frac{\mathrm{1}}{\mathrm{4}!}} \mathrm{5}^{\frac{\mathrm{1}}{\mathrm{5}!}} \mathrm{6}^{\frac{\mathrm{1}}{\mathrm{6}!}} \mathrm{7}^{\frac{\mathrm{1}}{\mathrm{7}!}} \mathrm{8}^{\frac{\mathrm{1}}{\mathrm{8}!}} \mathrm{9}^{\frac{\mathrm{1}}{\mathrm{9}!}} \mathrm{10}^{\frac{\mathrm{1}}{\mathrm{10}!}} = \\ $$$$=\mathrm{2}^{\frac{\mathrm{2122111}}{\mathrm{3628800}}} \mathrm{3}^{\frac{\mathrm{20329}}{\mathrm{120960}}} \mathrm{5}^{\frac{\mathrm{30241}}{\mathrm{3628800}}} \mathrm{7}^{\frac{\mathrm{1}}{\mathrm{5040}}} \\ $$$$\mathrm{this}\:\mathrm{looks}\:\mathrm{nasty}…\:\mathrm{we}\:\mathrm{should}\:\mathrm{hope}\:\mathrm{for}\:\mathrm{another} \\ $$$$\mathrm{Riemann}… \\ $$
Commented by TawaTawa last updated on 29/Jan/20
$$\mathrm{Sir}\:\mathrm{mrW},\:\mathrm{come}\:\mathrm{and}\:\mathrm{help}\:\mathrm{me}\:\mathrm{in}\:\:\mathrm{Q79943} \\ $$
Commented by mr W last updated on 29/Jan/20
Commented by mr W last updated on 29/Jan/20
$${i}\:{think}\:{P}_{{n}} =\sqrt{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{4}}]{\mathrm{4}…\sqrt[{{n}}]{{n}}}}}\:{converges}. \\ $$$${how}\:{can}\:{we}\:{find}\:\left({if}\:{exists}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{P}_{{n}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\sqrt{\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{3}\sqrt[{\mathrm{4}}]{\mathrm{4}…\sqrt[{{n}}]{{n}}}}}=? \\ $$
Commented by mr W last updated on 29/Jan/20
$${thanks}\:{sir}! \\ $$$${is}\:{any}\:“{exact}''\:{solution}\:{possible}? \\ $$