Let $a_n=\sum_{k=0}^{n-1}\frac{(-1)^{k}}{k!}$. Prove that this sequence is Cauchy.
I know that this is the partial sum of the series $\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}$ wich converges, but I don't know how to prove this, without using this fact, I mean, prove it by definition $|a_n-a_m|< \varepsilon$.
Let $m\geq n$. Then you have
$|a_m-a_n|=|\sum_{k=0}^{m-1}\frac{(-1)^{k}}{k!}-\sum_{k=0}^{n-1}\frac{(-1)^{k}}{k!}|=|\sum_{k=0}^{n-1}\frac{(-1)^{k}}{k!}+\sum_{k=n}^{m-1}\frac{(-1)^{k}}{k!}-\sum_{k=0}^{n-1}\frac{(-1)^{k}}{k!}|=|\sum_{k=n}^{m-1}\frac{(-1)^{k}}{k!}|\leq\sum_{k=n}^{m-1}\frac{1}{k!}<\epsilon$