- Montrer que : $\forall n \in \mathbb{N}, \quad u_{n+1} = u_n - \frac{1}{(n+1)!}$
- En déduire que : $\forall n \in \mathbb{N}, \quad u_n = e - \sum_{k=0}^n \frac{1}{k!}$
Pour tout $n \in \mathbb{N}$ on pose : $u_n = \frac{1}{n!} \int_0^1 (1-x)^n e^x \,dx$
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