Question

Lim n->infinit (n^rad n!)/n

Answer 1

Explicație pas cu pas:

Vom folosi criteriul Cauchy-d'Alembert:

$\displaystyle \lim_{ x \to \infty } \frac{ \sqrt[n]{n!} }{n }=\displaystyle \lim_{ x \to \infty } \bigg(\frac{ {n!} }{ n^n }\bigg)^{\frac{1}{n} }=\displaystyle \lim_{ x \to \infty } \sqrt[n]{\frac{n!}{n^n} }$

$Fie \ sirul \ (a_n), \ a_n=\frac{n!}{n^n} > 0 \ \forall n \in \mathbb{N^*}\\\exists \displaystyle \lim_{ n \to \infty } \frac{a_{n+1}}{a_n} =\displaystyle \lim_{ n \to \infty } \frac{(n+1)!}{(n+1)^{(n+1)}}\cdot \frac{n^n}{n!} =\displaystyle \lim_{ n \to \infty } \frac{n!\cdot(n+1)}{(n+1)^{n+1}} \cdot\frac{n^n}{n!} = \displaystyle \lim_{ n \to \infty } \frac{n^n}{(n+1)^{n}} =$

$\displaystyle \lim_{ n \to \infty } \bigg(\frac{n}{n+1} \bigg)^{n}=e^{-1}=\frac{1}{e} \implies \boxed{\displaystyle \lim_{ n \to \infty } \sqrt[n]{a_n} = \frac{1}{e}}$