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Question Number 79359 by ahmadshahhimat775@gmail.com last updated on 24/Jan/20

Commented by mathmax by abdo last updated on 24/Jan/20
$let A_n =(((n!)^(1/n) )/n) we have n! ∼n^n e^(−n) (√(2πn))( stirling formulae) ⇒ (n!)^(1/n) ∼n e^(−1) ((√(2πn)))^(1/n) ⇒ A_n ∼e^(−1) (2πn)^(1/(2n)) we have (2πn)^(1/(2n)) =e^((1/(2n))ln(2πn)) =e^((1/(2n)){ln(2π)+ln(n)}) =e^(((ln(2π))/(2n))+((ln(n))/(2n))) →e^0 =1 ⇒ lim_(n→+∞) A_n =(1/e)$
$l e t A_{n} = \frac{{(n!)}^{\frac{1}{n}}}{n} w e h a v e n! \sim n^{n} e^{- n} \sqrt{2 π n} (s t i r l i n g f o r m u l a e) \Rightarrow$ ${(n!)}^{\frac{1}{n}} \sim n e^{- 1} {(\sqrt{2 π n})}^{\frac{1}{n}} \Rightarrow A_{n} \sim e^{- 1} {(2 π n)}^{\frac{1}{2 n}} w e h a v e$ ${(2 π n)}^{\frac{1}{2 n}} = e^{\frac{1}{2 n} l n (2 π n)} = e^{\frac{1}{2 n} {l n (2 π) + l n (n)}} = e^{\frac{l n (2 π)}{2 n} + \frac{l n (n)}{2 n}} \to e^{0} = 1 \Rightarrow$ ${l i m}_{n \to + \infty} A_{n} = \frac{1}{e}$
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