The sum of the series $\sum_{n=0}^{\infty}\frac{\epsilon_n}{n!}$ is an irrational number Let $\{\epsilon_n\}$ be a sequence where $\epsilon_n$ is either $ 1$ or $-1$. How could I Show
that the sum of the series 
$$\sum_{n=0}^{\infty}\frac{\epsilon_n}{n!}$$
is an irrational number. 

If there exist $p\in\mathbb{Z}$ and $q\in\mathbb{N}$, such that
$\frac{p}{q}=\sum_{n=0}^\infty\frac{\epsilon_n}{n!}$, then $q!\cdot\sum_{n=q+1}^\infty\frac{\epsilon_n}{n!}$ must be an integer. However, 
$$|q!\cdot\sum_{n=q+1}^\infty\frac{\epsilon_n}{n!}-\frac{\epsilon_{q+1}}{q+1}|\le\sum_{n=q+2}\frac{q!}{n!}<\frac{1}{(q+1)(q+2)}\cdot\sum_{m=0}\frac{1}{2^m}=\frac{2}{(q+1)(q+2)},$$
which implies that
$$0<\frac{1}{q+1}-\frac{2}{(q+1)(q+2)}\le |q!\cdot\sum_{n=q+1}^\infty\frac{\epsilon_n}{n!}|\le \frac{1}{q+1}+\frac{2}{(q+1)(q+2)}<1,$$
a contradiction.
