prove-lim-n-n-n-x-n-x-1-

Question Number 198855 by mokys last updated on 25/Oct/23

p r o v e : \underset{n \to \infty}{l i m} \frac{n!}{n^{x} (n - x)!} = 1

Answered by witcher3 last updated on 25/Oct/23

\frac{Γ (n + 1)}{Γ (n + 1 - x) n^{x}}

Γ (z) = \sqrt{2 π} . z^{z - \frac{1}{2}} e^{- z}

\frac{Γ (n + 1)}{n^{x} (n - x)!} \sim \frac{\sqrt{2 π} {(n + 1)}^{n + \frac{1}{2}} e^{- (n + 1)}}{n^{x} \sqrt{2 π} . {(n + 1 - x)}^{n + \frac{1}{2} - x} e^{- (n + 1 - x)}}

= \lim_{n \to \infty} . \frac{e^{- x}}{n^{x}} . {(\frac{n + 1}{n + 1 - x})}^{n + \frac{1}{2}} . \frac{1}{{(n + 1 - x)}^{- x}}

{(\frac{n + 1}{n + 1 - x})}^{n + \frac{1}{2}} = e^{(n + \frac{1}{2}) \ln (1 + \frac{x}{n + 1 - x})}

= e^{(n + \frac{1}{2}) . (\frac{x}{n + 1 - x} - \frac{1}{2} {(\frac{x}{n + 1 - x})}^{2} + o (\frac{1}{n^{2}})}

\sim e^{x}

n^{x} . {(n + 1 - x)}^{- x} = {(\frac{n}{n + 1 - x})}^{x}

= {\underset{n \to \infty}{\lim e}}^{xln (1 + \frac{x - 1}{n + 1 - x})} \to 1

\frac{e^{- x}}{n^{x}} . {(\frac{n + 1}{n + 1 - x})}^{n + \frac{1}{2}} . \frac{1}{{(n + 1 - x)}^{- x}}

Missing \left or extra \right

\Rightarrow \lim_{n \to \infty} . \frac{Γ (n + 1)}{Γ (n + 1 - x) . n^{x}} = 1

Commented by mokys last updated on 25/Oct/23

t h a n k y o u s i r

Commented by witcher3 last updated on 27/Oct/23

Withe Pleasur