find-lim-n-n-2-n-1-

Question Number 29509 by abdo imad last updated on 09/Feb/18

f i n d {l i m}_{n \to + \infty} \frac{n!}{2^{n - 1}} .

Commented by abdo imad last updated on 11/Feb/18

l e t u s e s t i r l i n g f o r m u l a n! \sim n^{n} e^{- n} \sqrt{2 π n} \Rightarrow

\frac{n!}{2^{n - 1}} = n^{n} 2^{1 - n} e^{- n} n^{\frac{1}{2}} \sqrt{2 π} = e^{n l n (n)} e^{(1 - n) l n (2) - n + \frac{1}{2}} \sqrt{2 π}

= \sqrt{2 π} e^{n l n (n) + (1 - n) l n (2) - n + \frac{1}{2}} = \sqrt{2 π} e^{n (l n n + (\frac{1}{n} - 1) l n 2 - 1 + \frac{1}{2 n})}

a n d l i m \frac{n!}{2^{n - 1}} = {l i m}_{n \to \infty} \sqrt{2 π} e^{n (l n (n) - 1)} = + \infty .