Questions mathématiques diverses

SigMathS

Question 40:

\lim\limits_{n \to +\infty}\left({\dfrac{n!}{n^n}}\right)^{\dfrac{3n^3+4}{4n^4-1}}

$\lim\limits_{n \to +\infty}\left({\dfrac{n!}{n^n}}\right)^{\dfrac{3n^3+4}{4n^4-1}}$

SigMathS

Réponse 40:
On pose

u_n=\left({\dfrac{n!}{n^n}}\right)^{\dfrac{3n^3+4}{4n^4-1}}

\begin{align*} \ln(u_n)&=\dfrac{3n^3+4}{4n^4-1}\ln\left({\dfrac{1.2.3...n}{n.n.b...n}}\right)\\ &=\dfrac{3n^3+4}{4n^4-1}\sum\limits_{k=1}^{n}{\ln\left({\dfrac{k}{n}}\right)}\\ &=\dfrac{3n^4+4n}{4n^4-1}\left({\dfrac{1}{n}\sum\limits_{k=1}^{n}{\ln\left({\dfrac{k}{n}}\right)}}\right)\\ \lim\limits_{n \to +\infty}&\ln(u_n)=\dfrac{3}{4}\int_{0}^{1}{\ln x\;dx}=-\dfrac{3}{4} \end{align*}

Donc

\lim\limits_{n \to +\infty}u_n=e^{-\frac{3}{4}}

\int_{0}^{1}{\ln x\,dx}=\lim\limits_{\epsilon \to 0^+}\int_{\epsilon}^{1}{\ln x\,dx}

=\lim\limits_{\epsilon \to 0^+}\left[{x\ln x-x}\right]_\epsilon^1=-1