Convergence of the sequence $y_{n+1}=1-(n+1)y_n , y_0=(e-1)/e$ I would like to prove that the sequence $\{y_n\}_n$ defined recursively by $y_0=(e-1)/e$ and
$$y_{n+1}=1-(n+1)y_n , \quad n\geq 0$$
is convergent (if this is done then it clearly converges to $0$).
I tried to prove that $\{y_n\}_n$ is decreasing and bounded below by $0$ using mathematical induction, but the inequalities provided in the induction hypothesis are reversed and thus it's not possible to continue.
Could anyone please provide a hint/reference for this problem?
 A: I would find the general term of the sequence as follows :
Let $ n,k\in\mathbb{N}^{*} $, we have the following : \begin{aligned}x_{k}&=1-k x_{k-1}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(-1\right)^{k}\frac{x_{k}}{k!}&=\frac{\left(-1\right)^{k}}{k!}+\left(-1\right)^{k-1}\frac{x_{k-1}}{\left(k-1\right)!}\\ \iff \sum_{k=1}^{n}{\left(\left(-1\right)^{k}\frac{x_{k}}{k!}-\left(-1\right)^{k-1}\frac{x_{k-1}}{\left(k-1\right)!}\right)}&=\sum_{k=1}^{n}{\frac{\left(-1\right)^{k}}{k!}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(-1\right)^{n}\frac{x_{n}}{n!}-\frac{\mathrm{e-1}}{\mathrm{e}}&=\sum_{k=1}^{n}{\frac{\left(-1\right)^{k}}{k!}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(-1\right)^{n}\frac{x_{n}}{n!}&=\sum_{k=0}^{n}{\frac{\left(-1\right)^{k}}{k!}}-\frac{1}{\mathrm{e}}\end{aligned}
Using the fact that $ \sum\limits_{k=0}^{n}{\frac{\left(-1\right)^{k}}{k!}}-\frac{1}{\mathrm{e}}=\frac{\left(-1\right)^{n}}{n!}\int_{0}^{1}{\left(1-x\right)^{n}\,\mathrm{e}^{-x}\,\mathrm{d}x}$, we get that : $$\fbox{$\begin{array}{rcl} \left(\forall n\in\mathbb{N} \right),\ x_{n}=\displaystyle\int_{0}^{1}{\left(1-x\right)^{n}\,\mathrm{e}^{-x}\,\mathrm{d}x}\end{array}$} $$
Now for the limit, we could do something like : $$ \left\vert\int_{0}^{1}{\left(1-x\right)^{n}\,\mathrm{e}^{-x}\,\mathrm{d}x}\right\vert\leq\int_{0}^{1}{\left(1-x\right)^{n}\,\mathrm{d}x}=\frac{1}{n+1}\underset{n\to +\infty}{\longrightarrow}0 $$
Thus $ \left(x_{n}\right)_{n} $ converges, and $ x_{n}\underset{n\to +\infty}{\longrightarrow}0 $.
