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: 
The numbers you describe are in fact all irrational.

Given a fixed rational number $p/q$, any other rational number $a/b$ that approximates $p/q$ (but is not equal to $p/q$) can only do so to order $1/b$. This can be seen as follows:
$$ p/q - a/b = (pb - qa)/qb,$$
which is at least $(qb)^{-1}$ if it is not zero. So rationals cannot be approximated too well.

Now consider the series $$\sum x\_n/n!$$ and the partial sums $S\_j$. The partial sums $S\_k$ have denominator $k!$. Their error from the infinite sum is at most $\sum\_{n>k} 1/n!$ which (from standard upper bounds via geometric series) little oh of $1/k!$ (as in the standard proof for $e$). 

So basically, the problem is that the series converges "too rapidly" for the sum to be rational. A rational number cannot be approximated too well (well in the denominators) by other rationals.

Note that there are similar Diophantine approximation results for algebraic numbers, cf. [Roth's theorem](http://en.wikipedia.org/wiki/Thue%E2%80%93Siegel%E2%80%93Roth_theorem). Unfortunately that's not applicable here, though, since the approximations $S\_k$ are little oh of the denominator, not little oh of a power of the denominator.
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.