Rationality of series $\sum \frac{1}{n!}$

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.

Solution 1:

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. 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.

Solution 2:

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? Can you adapt it here?

Solution 3:

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.