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Last Update: 12/02/2024 07:52 AM
Current Deck: Sur papier::Maths::Suites
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Commit #225724
[latex]
Déjà :
\begin{equation*}
\begin{aligned}
u_n & = \ln \left(n + \sqrt{n^2+1}\right) \\
& = \ln \left(n + n \sqrt{1 + \frac{1}{n^2}}\right) \\
\end{aligned}
\end{equation*}
Lorsque $n \to +\infty$ :
\begin{equation*}
\sqrt{1 + \frac{1}{n^2}} \sim 1 + \frac{1}{2n^2}
\end{equation*}
Donc :
\begin{equation*}
n + n\sqrt{1 + \frac{1}{n^2}} \sim 2n + \frac{1}{2n}
\end{equation*}
Ainsi :
\begin{equation*}
\ln \left(n + n \sqrt{1 + \frac{1}{n^2}}\right) \sim \ln(2n)
\end{equation*}
Ccl. $u_n \sim\ln(2n)\ln(2n)$
[/latex]
Déjà :
\begin{equation*}
\begin{aligned}
u_n & = \ln \left(n + \sqrt{n^2+1}\right) \\
& = \ln \left(n + n \sqrt{1 + \frac{1}{n^2}}\right) \\
\end{aligned}
\end{equation*}
Lorsque $n \to +\infty$ :
\begin{equation*}
\sqrt{1 + \frac{1}{n^2}} \sim 1 + \frac{1}{2n^2}
\end{equation*}
Donc :
\begin{equation*}
n + n\sqrt{1 + \frac{1}{n^2}} \sim 2n + \frac{1}{2n}
\end{equation*}
Ainsi :
\begin{equation*}
\ln \left(n + n \sqrt{1 + \frac{1}{n^2}}\right) \sim \ln(2n)
\end{equation*}
Ccl. $u_n \sim
[/latex]