exercicio resolvido

por **adauto martins** » Seg Abr 12, 2021 15:59

(ITA-1952) calcular o
$\lim_{n\rightarrow\infty}\sqrt[n]{n!}/n$

por **adauto martins** » Seg Abr 12, 2021 17:04

soluçao
precisarei de dois argumentos para resolver esse exercicio

o limite fundamental

$\lim_{n\rightarrow\infty}(1+(1/n))^n=\lim_{y\rightarrow 0}(1+y)^{1/y}=e$

e o limite,que é consequencia do limite fundamental apresentado

$\lim_{y\rightarrow 0}(1+ny)^{1/y}=e^n$

de fato,

$(1+ny)^{1/y}=((1+ny)^{1/y})^{n}$

façamos

$z=ny\Rightarrow \Rightarrow ((1+z)^{1/z})^{n}$

logo

$\lim_{z\rightarrow 0}( ((1+z)^{1/z})^{n})=(\lim_{z\rightarrow0}(1+z)^{1/z})^{n}=e^{n}$

voltemos a questao

$L=\sqrt[n]{n!}/n=\sqrt[n]{n!/n^{n}}=((n.(n-1)....2.1)/n^{n})^{1/n} =((n/n).(n-1)/n....(2/n).(1/n))^{1/n} =(1-1/n)^{1/n}.(1-2/n)^{1/n} ....(1-(n-2)/n)^{1/n}.(1-(n-1)/n)^{1/n}$

façamos

$y=-(1/n)...n\rightarrow\infty...y\rightarrow0$

entao

$\sqrt[n]{n!}/n=(1+y)^{-1/y}.(1+2y)^{-2/y}....(1+(n-1)y)^{-(n-1)/y} \lim_{y\rightarrow0}((1+y)^{-1/y}.(1+2y)^{-2/y}....(1+(n-1)y)^{-(n-1)/y}) (\lim_{y\rightarrow0}(1+y)^{-1/y})....(\lim_{y\rightarrow0}(1+(n-1)y)^{-(n-1)/y})$

$=e^{-1}.e^{-2}....e^{(n-1)}=(e^{1+2+...+(n-1)}) ^{-1} =(e^{((n-1).(n-2))/2})^{-1}=(e^{((n^2-3.n+2)/2)})^{-1} =(e^{n^2-(3/2)n+1}) =(e^{n^2/2)})^{-1}.(e^{-3n/2})^{-1}.e^{-1} =(e^{(2/n^2)-2/(3n)+1)} =e^{2/n^2}.e^{(-2/3n)}.e^{-1} n\rightarrow \infty\Rightarrow L=1.1.e^{-1}=1/e...$

exercicio resolvido

exercicio resolvido

exercicio resolvido

Re: exercicio resolvido

Quem está online