How to prove that a sequence where each term is a summation is Cauchy

I'm given that $s_n=1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}$, or, $s_n=\sum_{i=1}^{n}\frac{1}{i!}$.

I'm supposed to prove that this sequence is Cauchy. So I do the usual setup, where we assume $\epsilon >0$ and for $k\in\mathbb{N}$. Then I come across the problem. We have

$$\left|s_{n+k}-s_n\right|=\left|\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\cdots+\frac{1}{(n+k)!}-1-\cdots-\frac{1}{n!}\right|$$

My problem is two-fold:

1. This doesn't feel algebraically correct. In the book, there is an example which uses $s_n=1+\frac{1}{2^2}+\cdots+\frac{1}{n^2}$, and during the proof $\left|s_{n+k}-s_n\right|=\frac{1}{(n+1)^2}+\cdots+\frac{1}{(n+k)^2}$, which I thought was just $s_{n+1}$.
2. How do you go about, in general, proving that a sequence made up of summations is Cauchy?
• The $|s_{n+k}-s_n|$ formula is wrong – user175968 Oct 26 '15 at 6:21
• Jaska"s formula for |s_{n+k}-s_n| is wrong.The book's formula is right. – DanielWainfleet Oct 26 '15 at 7:22

By completeness of the reals, a sequence is convergent iff it is Cauchy. Note that $$\frac{1}{i!} < \frac{1}{2^{i}} = \frac{1}{\exp ( i\log 2)} < \frac{1}{i^{2}}$$ for large $i$; hence by the comparison test $(s_{n})$ converges and then $(s_{n})$ is Cauchy.
Regarding your first question, we have $$|s_{n+k} - s_{n}| = \Big| \frac{1}{(n+k)!} + \cdots + \frac{1}{(n+1)!} \Big|,$$ rather than yours.