For all $n\in\mathbb{N}$, we define: $$x_{n}=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{(-1)^{n+1}}{n!}$$
I want to show that $\forall\space m,n\in\mathbb{N}$ such that $m\ge n$:
$$|x_{m}-x_{n}|<\frac{1}{nn!}$$
I managed to show $$|x_{m}-x_{n}|\le \frac{1}{(n+1)!}+...+\frac{1}{m!}$$ but don't know where to go from here.
If $$ a_1 \ge a_2 \ge a_3 \ge a_4 \ge \ldots \ge a_N \ge 0 $$ are decreasing, non-negative real numbers then the alternating sum $$ S = a_1 - a_2 + a_3 - a_4 + \ldots \pm a_N \\ $$ satisfies $$ a_1 - a_2 \le S \le a_1 \, , $$ as can be seen by grouping pairs of consecutive terms in two possible ways: $$ \begin{aligned} S &= a_1 - a_2 + (a_3 - a_4) + (a_5 - a_6) + \ldots \quad \ge a_1 - a_2 \, ,\\ S &= a_1 - (a_2 - a_3) - (a_4 - a_5) - \ldots \quad \le a_1 \, . \end{aligned} $$
Applying this to $$ (-1)^n (x_m - x_n) = \frac{1}{(n+1)!} - \frac{1}{(n+2)!} + \ldots \pm \frac{1}{m!} $$ gives $$ \frac{1}{(n+1)!} - \frac{1}{(n+2)!} \le |x_m-x_n|\le \frac{1}{(n+1)!}<\frac{1}{nn!} \, . $$