I have already proven that $S_n(x) :=\sum_{k=1}^n \frac{x^k}{k!}$ is convergent for all real $x>0$. For this I used the Monotone Convergence Theorem. Clearly this implies that $S_n(|x|)$ is convergent for all $x\in\mathbb{R}$.
Now I wish to prove that the sequence $(S_n(x))_{n\in\mathbb{N}}$ is a Cauchy sequence for all $x\in\mathbb{R}$. I have arrived at the following inequality (assuming WLOG $m>n$):
$$|S_m(x) - S_n(x)|\leq|S_m(|x|) - S_n(|x|)|.$$
The lecture notes say that this implies that $(S_n(x))_{n\in\mathbb{N}}$ is a Cauchy sequence. I cannot quite see why this is the case, because I can't see how it implies there exists an $N$ such that with $m,n>N$, we have $|S_m(|x|) - S_n(|x|)|<\epsilon$. Pointers in the right direction would be appreciated.
Edit: To clarify, I have not yet shown that $S_n(x)$ is convergent for all $x$, only $x>0$. The lecture notes show that $S_n(x)$ is convergent for all $x$, by showing that $S_n(x)$ is Cauchy, and uses an already proven result that $S_n(|x|)$ is convergent to do this. I am asking for help in understanding how to show $S_n$ is Cauchy, in particular using this scheme.
You already know that $S_n(\lvert x\rvert)$ converges for all $x\in\mathbb R$. Therefore, it is a Cauchy sequence especially.