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Question Number 79932 by mr W last updated on 29/Jan/20

Commented by mr W last updated on 29/Jan/20

$[Q 79861 r e p o s t e d]$
	Answered by mr W last updated on 29/Jan/20

	$I t h i n k I^{'} v e f o u n d a n e a s y w a y t o p r o v e :$ ${l e t}^{'} s l o o k a t f (x) = \ln (x), w e s e e$ $f (x) = \ln x ⩽ \frac{x}{e} (‘ ‘ =^{″} i s o n l y a t x = e)$ $(s e e d i a g r a m b e l o w i f y o u h a v e d o u b t)$ $P_{n} = \sqrt{2 \sqrt[3]{3 \sqrt[4]{4 . . . \sqrt[n]{n}}}}$ $P_{n} = 2^{\frac{1}{2}} 3^{\frac{1}{2} \times \frac{1}{3}} 4^{\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}} . . . n^{\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4} \times . . . \times \frac{1}{n}}$ $P_{n} = 2^{\frac{1}{2!}} 3^{\frac{1}{3!}} 4^{\frac{1}{4!}} . . . n^{\frac{1}{n!}} = \underset{k = 2}{\prod^{n}} k^{\frac{1}{k!}}$ $\ln P_{n} = \underset{k = 2}{\sum^{n}} \frac{\ln k}{k!} < \underset{k = 2}{\sum^{n}} \frac{k}{e k!} = \frac{1}{e} \underset{k = 2}{\sum^{n}} \frac{1}{(k - 1)!} = \frac{1}{e} \underset{k = 1}{\sum^{n - 1}} \frac{1}{k!}$ $< \frac{1}{e} \underset{k = 1}{\sum^{\infty}} \frac{1}{k!} = \frac{1}{e} (\underset{k = 0}{\sum^{\infty}} \frac{1}{k!} - 1) = \frac{1}{e} (e - 1) = 1 - \frac{1}{e}$ $\Rightarrow P_{n} < e^{1 - \frac{1}{e}} = \frac{e}{\sqrt[e]{e}} \approx 1.8816 < 2 \Rightarrow p r o v e d$
	Commented by M±th+et£s last updated on 29/Jan/20

	$t h a n k y o u s i r g r e a t s o l u t i o n$
	Commented by MJS last updated on 29/Jan/20

	$\lim_{n \to \infty} P_{n} \approx 1.82902467956$
	Commented by MJS last updated on 29/Jan/20

	$P_{n} = 2^{\frac{1}{2!}} 3^{\frac{1}{3!}} 4^{\frac{1}{4!}} . . . n^{\frac{1}{n!}}$ $looking at the prime factors . . .$ $2^{\frac{1}{2!}} 3^{\frac{1}{3!}} 4^{\frac{1}{4!}} 5^{\frac{1}{5!}} 6^{\frac{1}{6!}} 7^{\frac{1}{7!}} 8^{\frac{1}{8!}} 9^{\frac{1}{9!}} 10^{\frac{1}{10!}} =$ $= 2^{\frac{2122111}{3628800}} 3^{\frac{20329}{120960}} 5^{\frac{30241}{3628800}} 7^{\frac{1}{5040}}$ $this looks nasty . . . we should hope for another$ $Riemann . . .$
	Commented by TawaTawa last updated on 29/Jan/20

	$Sir mrW, come and help me in Q 79943$
	Commented by mr W last updated on 29/Jan/20

	Commented by mr W last updated on 29/Jan/20

	$i t h i n k P_{n} = \sqrt{2 \sqrt[3]{3 \sqrt[4]{4 . . . \sqrt[n]{n}}}} c o n v e r g e s .$ $h o w c a n w e f i n d (i f e x i s t s)$ $\lim_{n \to \infty} P_{n} = \lim_{n \to \infty} \sqrt{2 \sqrt[3]{3 \sqrt[4]{4 . . . \sqrt[n]{n}}}} = ?$
	Commented by mr W last updated on 29/Jan/20

	$t h a n k s s i r!$ $i s a n y ‘ ‘ {e x a c t}^{″} s o l u t i o n p o s s i b l e ?$
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