The following series is an irrational number $(*)$ Let $\{ x_n \}$ be a sequence where $x_n$ is either $1$ or $-1$. Then 
$$ \sum_{n=0}^{\infty} \frac{ x_n}{n!} \; \; \; \text{is irrational}$$
This problem arise when I was trying to prove that $e$ is irrational. If the result above is true, then we can put $x_n = (-1)^n $ and hence 
$$ \frac{1}{e} = \sum \frac{(-1)^n}{n!} \; \; \; \text{is irrational} $$
which would imply that $e$ is irrational. 
So, my question is: Is $(*)$ true? 
 A: The "standard" proof of the irrationality of $e$ goes like this:
$e = 1 + 1 + \frac{1}{2!} + ... + \frac{1}{r!} + ...$
Assume to the contrary that $e$ is rational, i.e. that $e = \frac{p}{q}$, where $p$ and $q$ are coprime integers. Clearly, $q > 1$
Now multiply both sides by $q!$:
$p!(q-1)! = q! + \frac{q!}{1!} + ...\frac{q!}{q!} + \frac{q!}{(q+1)!} + ...$
The LHS is clearly an integer (call it $M$).
The first $q+1$ terms on the RHS, i.e. up to and including $\frac{q!}{q!}$ sum up to an integer. Call this partial sum $N$.
Call the remaining partial sum $S$.
By rearrangement, $S = M - N$. Hence $S$ is a positive integer. 
Also $S = \frac{1}{q+1} + \frac{1}{(q+1)(q+2)} + ...$
which can quite easily be shown to be strictly less than the geometric series:
$G = \frac{1}{q+1} + \frac{1}{(q+1)^2} + ...$
which sums to $\frac{1}{q} < 1$.
Therefore $0 < S < 1$.
Hence there is a positive integer strictly between $0$ and $1$, which is impossible, and we've arrived at a contradiction.
Hence our original assumption was false and $e$ is irrational.
