$x_n =\sum_{i=0}^{n}\frac{1}{i!}$ is Cauchy in $\mathbb{Q}$

Question

From Topology Without Tears:

Show that $x_n=\sum_{i=0}^{n}\frac{1}{i!}$ is a Cauchy sequence in $\mathbb{Q}$

First thing, I believe being convergent in $\mathbb{Q}$, means sequence and the converging point must also belong to $\mathbb{Q}$. What does it mean sequence is Cauchy in $\mathbb{Q}$?Does it just mean sequence is entirely in $\mathbb{Q}?$

I was trying to show Cauchy: let $n>m$ \begin{equation} \begin{split} |x_n-x_m|&=|\sum_{i=0}^{n}\frac{1}{i!}-\sum_{i=0}^{m}\frac{1}{i!}|\\ &=|\sum_{i=m+1}^{n}\frac{1}{i!}|\\ &=|\frac{1}{m+1!}+...+\frac{1}{n!}|\\ & \leq \frac{1}{m+1!}+...+\frac{1}{n!}\\ &<\frac{1}{m}+...\frac{1}{n} \end{split} \end{equation} Since the number of terms are $n-m $.I was thinking to take each term $<\frac{\epsilon}{n-m}$, to get $n_0\in \mathbb{N},\ such\ that\ |x_n-x_m|<\epsilon,\forall n,m>n_0$.

But I think this will be wrong and looks odd,saying choosing $m>\frac{\epsilon}{n-m}$

How should I show this is Cauchy?

Answer 1

$(m+1)!$ is a product of $m+1$ numbers all except one of which is $2$ or more. So $(m+1)!\geq 2^{m}$. Similarly, $(m+2)!\geq 2^{m+1}$ and so on. So $\frac 1 {m+1}!+..+\frac 1 {n!} \leq \frac 1 {2^{m}}+\frac 1 {2^{m+1}}+...$. Can you write down this geometric sum and show that it is less than $\epsilon$ if $m$ is large enough?