Question: There is a famous literature on $\displaystyle e = \sum\limits\_{n=0}^{\infty} \frac{1}{n!}$. We know that $e$ is irrational as well as transcendental. 

Question is: For each $x\_{n}$, sequence of $\pm{1}'s$, let $$f(x\_{n}) = \sum\limits\_{n=0}^{\infty} \frac{x\_{n}}{n!}$$ for which sequences $(x\_n)$ is $f(x\_{n})$ rational?

Also can we write $e$ as $\displaystyle e = \sum\limits\_{n=1}^{\infty} \frac{x\_{n}}{n}$ where $x\_{n}$ is a sequence as above.

Answer: 
**Suggestion for the first question** [I have not checked all the details myself]: Have you tried to modify the proof of the irrationality of $e$ to deal with your similar-looking series? 

Specifically, referring to the standard proof given in 

<http://en.wikipedia.org/wiki/Proof_that_e_is_irrational>

which steps require modification if $\sum\_{n=0}^{\infty} \frac{1}{n!}$ is replaced by 
$\sum\_{n=0}^{\infty} \frac{x\_n}{n!}$ for $x\_n \in \{ \pm 1 \}$?

**Hint for the second question**: are you familiar with the proof of the [Riemann Rearrangement Theorem](http://en.wikipedia.org/wiki/Riemann_rearrangement_theorem)? Can you adapt it here?
The answer by Akhil Mathew is incomplete. It remains to check that $\sum\_{n>k} x\_n/n!\neq 0$ for infinitely many $k$. This is not obviously true here, it might even be false for some sequence $(x\_n)$. Akhil Mathew's argument could be used for the series $\sum\_{n\ge 0} n/(n+1)!$ which is similar to $\sum\_{n\ge 0} 1/n!$. But its sum is equal to 1.