Please give hint to find the sum of this series: $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n!}$

Question

can anyone give me some hint how to find this series's sum? $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n!}$$ I know that by the ratio test, I can find that it is convergent. But in order to find its sum, I may need some analysis technique which I don't have. thanks.

Answer 1

One may recall that $$ \sum_{n=0}^{\infty}\frac{z^{n}}{n!}=e^z,\quad z \in \mathbb{C}. $$

Answer 2

$$e^{-1}=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{n!}=1+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n!}=1-\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n!}$$ so $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n!}=1-e^{-1}=\frac{e-1}{e}$$