Pour \(n\geq 5\), on a

\[u_n=1+\frac{1}{n}+\frac{1}{n(n-1)}+\frac{1}{n(n-1)(n-2)}+\frac{1}{n(n-1)(n-2)(n-3)}+\sum_{k=0}^{n-5}\frac{1}{n(n-1)...(k+1)}.\] Ensuite,

\[0\leq n^3\sum_{k=0}^{n-5}\frac{1}{n(n-1)...(k+1)}\leq n^3(n-4)\frac{1}{n(n-1)(n-2)(n-3)(n-4)}\underset{n\rightarrow+\infty}{\sim}\frac{1}{n}\underset{n\rightarrow+\infty}{\rightarrow}0.\] Donc, \(\sum_{k=0}^{n-5}\frac{1}{n(n-1)...(k+1)}\underset{n\rightarrow+\infty}{=}o\left(\frac{1}{n^3}\right)\) et de même \(\frac{1}{n(n-1)(n-2)(n-3)}\underset{n\rightarrow+\infty}{=} o\left(\frac{1}{n^3}\right)\). Il reste

\[\begin{align*}\ensuremath u_n&\underset{n\rightarrow+\infty}{=}1+\frac{1}{n}+\frac{1}{n^2}\left(1-\frac{1}{n}\right)^{-1}+\frac{1}{n^3}+o\left(\frac{1}{n^3}\right)=1+\frac{1}{n} +\frac{1}{n^2}\left(1+\frac{1}{n}\right)+\frac{1}{n^3}+o\left(\frac{1}{n^3}\right)\\ &=1+\frac{1}{n}+\frac{1}{n^2}+\frac{2}{n^3}+o\left(\frac{1}{n^3}\right). \end{align*}\]

\(\frac{1}{n!}\sum_{k=0}^{n}k!\underset{n\rightarrow+\infty}{=}1+\frac{1}{n}+\frac{1}{n^2}+\frac{2}{n^3}+o\left(\frac{1}{n^3}\right)\).

exo7 5432