

\begin{enumerate}
\item Pour tout entier naturel $k$,
\begin{align*}
\dfrac{k}{(k+1)!} & =\dfrac{k+1}{(k+1)!} - \dfrac{1}{(k+1)!} = \dfrac{\cancel{(k+1)}}{k!\times\cancel{(k+1)}}-\dfrac{1}{(k+1)!} = \dfrac{1}{k!}-\dfrac{1}{(k+1)!}.
\end{align*}

\item $\begin{aligned}[t]
S_n & = \sum_{k=1}^n \dfrac{k}{(k+1)!} = \sum_{k=1}^n \left( \dfrac{1}{k!}-\dfrac{1}{(k+1)!} \right)\\
& = \left( \dfrac{1}{1!} - \dfrac{1}{2!}\right) + \left( \dfrac{1}{2!} - \dfrac{1}{3!}\right) + \left( \dfrac{1}{3!} - \dfrac{1}{4!}\right) + \cdots + \left( \dfrac{1}{n!} - \dfrac{1}{(n+1)!}\right)\\
& = \dfrac{1}{1!} - \cancel{\dfrac{1}{2!}}  +  \cancel{\dfrac{1}{2!}} - \cancel{\dfrac{1}{3!}}  +  \cancel{\dfrac{1}{3!}} - \cancel{\dfrac{1}{4!}}  + \cdots +  \cancel{\dfrac{1}{n!}} - \dfrac{1}{(n+1)!} \\
& = 1 - \dfrac{1}{(n+1)!}.
\end{aligned}$


\item $\limite{n}{+\infty}{(n+1)!}=+\infty$ donc $\limite{n}{+\infty}{\dfrac{1}{(n+1)!}}=0$.

Ainsi, $\boxed{\limite{n}{+\infty}{S_n}=1}$.
\end{enumerate}
