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\begin{document}

%%%%%%%%%%b{gedeonInfo}%%%%%%%%%
%%--Document N� 1315033419238--
%%--Corrig� du document N� 1315033695949--
%%%%%%%%%%e{gedeonInfo}%%%%%%%%%


\begin{claim}
$\vspace*{-0.35cm}$

\begin{enumerate}
\item $u_{n}=1+\tfrac{1}{1!}+\tfrac{1}{2!}+\cdots+\tfrac{1}{n!}$ et
$v_{n}=u_{n}+\tfrac{1}{n!}.$

Pour d\'{e}montrer que $(u_{n})$ et $\left(  v_{n}\right)  $ sont adjacentes
il suffit de d\'{e}montrer $\underset{n\rightarrow+\infty}{\lim}\left(
u_{n}-v_{n}\right)  =0$ et que $\left(  u_{n}\right)  $ et $\left(
v_{n}\right)  $ ont des variations contraires.

\begin{description}
\item[D\'{e}montrons que $\underset{n\rightarrow+\infty}{\lim}(u_{n}%
-v_{n})=0:$] on a: \fbox{$u_{n}-v_{n}=-\tfrac{1}{n!}\overset{+\infty
}{\rightarrow}0.$}

\item[D\'{e}montrons que $(u_{n})_{n\in{%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}}$ est croissante:] $\forall n\in{%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}$ on a:%
\begin{align*}
u_{n+1}-u_{n}  & =1+\tfrac{1}{1!}+\tfrac{1}{2!}+\cdots+\tfrac{1}{n!}+\tfrac
{1}{(n+1)!}-(1+\tfrac{1}{1!}+\tfrac{1}{2!}+\cdots+\tfrac{1}{n!})\\
& =\tfrac{1}{(n+1)!}>0
\end{align*}


\item donc $u_{n+1}>u_{n},$ ie: \fbox{$\left(  u_{n}\right)  _{n\in{%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}}$ est croissante.}

\item[D\'{e}montrons que $\left(  v_{n}\right)  _{n\in{%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}}$ est d\'{e}croissante:] $\forall n\in{%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}$ on a:%
\begin{align*}
v_{n+1}-v_{n}  & =u_{n+1}+\tfrac{1}{(n+1)!}-u_{n}-\tfrac{1}{n!}=u_{n+1}%
-u_{n}+\tfrac{1}{(n+1)!}-\tfrac{1}{n!}\\
& =\tfrac{1}{(n+1)!}+\tfrac{1}{(n+1)!}-\tfrac{1}{n!}\\
& =\tfrac{2}{(n+1)!}-\tfrac{n+1}{(n+1)!}=\tfrac{-(n+1)}{(n+1)!}<0
\end{align*}


\item donc $v_{n+1}<v_{n},$ ie: \fbox{$\left(  v_{n}\right)  _{n\in{%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}}$ est d\'{e}croissante.}
\end{description}

\item $u_{7}=1+\tfrac{1}{1!}+\tfrac{1}{2!}+\tfrac{1}{3!}+\tfrac{1}{4!}%
+\tfrac{1}{5!}+\tfrac{1}{6!}+\tfrac{1}{7!}=\allowbreak\dfrac{685}{252}%
\simeq\allowbreak2.\,\allowbreak7183$ \`{a} $10^{-4}$ pr\`{e}s.

$v_{7}=1+\tfrac{1}{1!}+\tfrac{1}{2!}+\tfrac{1}{3!}+\tfrac{1}{4!}+\tfrac{1}%
{5!}+\tfrac{1}{6!}+\tfrac{1}{7!}+\tfrac{1}{7!}=\allowbreak\dfrac{4567}%
{1680}=\allowbreak2.\,\allowbreak718\,5$ \`{a} $10^{-4}$ pr\`{e}s.

\begin{enumerate}
\item D'o\`{u}, comme $u_{7}<e<v_{7},\quad$\fbox{$e\simeq2.718$ \`{a}
$10^{-3}$ pr\`{e}s.}
\end{enumerate}

\item Suppons qu'il existe deux entiers strictement positifs $p$ et $q$ tel
que $e=\tfrac{p}{q}\quad($hypoth\`{e}se$\mathcal{H}).$

Comme $\left(  u_{n}\right)  $ est \textbf{strictement croissante} avec
$\underset{n\rightarrow+\infty}{\lim}u_{n}=e$, on a: $\forall n\in{%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
,u}_{n}<e$ et en particulier $u_{q}<e.$

\begin{enumerate}
\item De m\^{e}me, comme $\left(  v_{n}\right)  $ est \textbf{strictement
d\'{e}croissante} avec $\underset{n\rightarrow+\infty}{\lim}v_{n}=e$, on a:
$\forall n\in{%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
,v}_{n}>e$ et en particulier $v_{q}>e.$

Or:%
\begin{align*}
u_{q}  & <e<v_{q}\\
& \Leftrightarrow1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots+\dfrac{1}{q!}<\dfrac
{p}{q}<1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots+\dfrac{1}{q!}+\dfrac{1}{q!}\\
& \Leftrightarrow q!+\dfrac{q!}{1!}+\dfrac{q!}{2!}+\cdots+\dfrac{q!}%
{q!}<\dfrac{p}{q}\times q!<q!+\dfrac{q!}{1!}+\dfrac{q!}{2!}+\cdots+\dfrac
{q!}{q!}+\dfrac{q!}{q!}\\
& \Leftrightarrow m<p\times(q-1)!<m+\dfrac{q!}{q!}\quad\text{o\`{u}
}m=q!+\dfrac{q!}{1!}+\dfrac{q!}{2!}+\cdots+\dfrac{q!}{q!}\in{%
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
}\\
& \Leftrightarrow m<p\times(q-1)!<m+1\quad\text{\textbf{ce qui est impossible
car }}\mathbf{m}\text{\textbf{\ et }}\mathbf{p(q-1)!}\text{\textbf{\ sont des
entiers.}}%
\end{align*}


Donc l'hypoth\`{e}se $\mathcal{H}$ est absurde, \fbox{ce qui prouve que $e$
est irrationnel.}
\end{enumerate}
\end{enumerate}
\end{claim}


\end{document}