How to prove that $x_n=1+\frac{1}{2!}+\cdots+\frac{1}{n!}$ is a Cauchy sequence?

Let $(x_n)_{n\in\mathbb{N}}$ be a real sequence given by $$x_n=1+\frac{1}{2!}+\cdots+\frac{1}{n!}$$ I would like to prove that, for each $\varepsilon>0$, there exists $n_0\in\mathbb{N}$ such that if $n>m>n_0$ then $$|x_n-x_m|=\frac{1}{(m+1)!}+\cdots+\frac{1}{n!}<\varepsilon$$ In other words, I want to prove that $(x_n)$ is a Cauchy sequence. I know this sequence is a Cauchy sequence because it's convergent, but I want to prove it by definition of Cauchy sequence.

Thanks.

Solution 1:

We have:

$$|x_n-x_m|=\frac{1}{(m+1)!}+\cdots+\frac{1}{n!}\leq \sum_{k=m}^\infty \frac{1}{2^k}=\frac{1}{2^{m-1}}$$ and $\frac{1}{2^{m-1}}$ can be made as small as we want.

Solution 2:

Hint: $\frac{1}{(m+1)!}+\ldots + \frac{1}{(m+k)!}< \frac{1}{m!}(\frac{1}{m}+\frac{1}{m^2}+\ldots+\frac{1}{m^k})$