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Soient \(\alpha, \beta, a\in \Bbb R\) tq \(\alpha, \beta\gt 0\) et a>1
Alors :- $$\underset{n\longrightarrow\infty}\lim\frac{n^\alpha}{a^n}={{0}}$$
- $$\underset{n\longrightarrow\infty}\lim\frac {(ln (n))^\beta}{n^\alpha}={{0}}$$
- $$\underset{n\longrightarrow\infty}\lim\frac{(ln(n))^\beta}{a^n}={{0}}$$
\(\longrightarrow\) preuve de 1:
Posons \(U_n=\frac{n^\alpha}{a^n}\gt 0\)
Et \(V_n=U_n^\frac1\alpha=e^{\frac 1\alpha ln(U_n)}\)
On a: $$V_n=\frac {n}{a^{\frac {n}{\alpha} } }=\frac{n}{(a^{\frac{1}{\alpha} })^n }=\frac {n}{((b+1)^n}$$
Comme \(a\gt 1\) \(\implies\) \(a^{\frac {1}{\alpha} }\gt 1\) \(\implies\) \(b:=a^{\frac {1}{\alpha} }-1\gt 0\)
D'où $$V_n = \frac {n}{1+\begin{pmatrix} n\\ 1\end{pmatrix}b+....+\begin{pmatrix} n\\ k\end{pmatrix}b^k+....+\begin{pmatrix} n\\ n\end{pmatrix}b^n}$$
Binome N:
\(0\leq V_n \leq \frac {n}{\begin{pmatrix} n\\ 2\end{pmatrix}b^2}=\frac{1}{\frac{(n-1)}{2} b^2}\)
Donc
$$0\leq V_n \leq \frac{2}{(n-1) b^2}, n\geq 2$$
Donc \(V_n\) converge vers 0
On a \(U_n =V_n^\alpha\)
D'où: \(U_n\) tend vers 0
