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. 
 A: 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.
A: The proof is by contradiction. Define:
$$
S = \sum_{k \ge 0} \frac{\epsilon_k}{n!}
$$
Assume $S$ is rational,
i.e. there are $u \in \mathbb{Z}$, $v \in \mathbb{N}$ such that $S = u / v$.
Pick $b > v$, so that $b \ge 2$.
Then $b! S$ is an integer, i.e.:
$$
S = \sum_{0 \le k \le b} \frac{b! \epsilon_k}{k!} 
     + \sum_{k > b} \frac{b! \epsilon_k}{k!}
$$
The first sum is an integer, so the second sum has to be an integer too. Now:
$$
\frac{b! \epsilon_k}{k!} 
   = \frac{\epsilon_k}{(b + 1) (b + 2) \ldots k}
$$
But:
$$
\frac{1}{(b + 1) (b + 2) \ldots k}
 < \frac{1}{b^{k - b}}
$$
By the triangular inequality:
$$
\left\rvert \sum_{k \ge b + 1} \frac{b! \epsilon_k}{k!} \right\rvert
 \le \sum_{k \ge b + 1} \frac{b!}{k!}
 < \sum_{k \ge b + 1} b^{-k}
 = b^{- b - 1} \sum_{k \ge 0} b^{-k}
 = b^{- b - 1} \frac{1}{1 - 1 / b}
 = \frac{1}{b^{b +1} (b - 1)}
 < \frac{1}{b^{b + 1}}
$$
So:
$$
1 = \left\lvert \frac{b! \epsilon_b}{b!} \right\rvert
  > \left\rvert \sum_{k \ge b + 1} \frac{b! \epsilon_k}{k!} \right\rvert
$$
and the "leftover sum" can never be 0, so it isn't an integer.
