# 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.

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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.

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@Akhil: Excellent answer. –  anonymous Aug 15 '10 at 7:39
@Akhil. I don't understand what do you mean with your statement "a rational number $p/q$ can only be approximated to order $1/q$". :-? So, the rational number $0$ is approximated by the sequence $1/n$ to order...? –  Agustí Roig Aug 15 '10 at 13:21
@Agusti: The rational number zero is approximated by $1/n$ to the order $1/n$, Note that this tends to zero at the same order as the denominator (in particular, precisely as the denominator, but this is special). The problem is that in the above case, the sequence tends to zero too fast relative to the denominators. –  Akhil Mathew Aug 15 '10 at 15:17
@Akhil. I'm sorry, but I don't understand. In your first answer, you say that $p/q$ can be approximated by $a/b$ only to order $1/q$. Here, you say that $0$ can be approximated by $1/n$ only to order $1/n$. Maybe I'm missing something, but I'd say that you are changing the roles of $p/q$ and $0$ and those of $a/b$ and $1/n$ in your two answers. I mean, according to your first definition, I would say that, since $0 = 0/n$, for all non-negative integer numbers $n$, then $0$ -and hence any rational number- can be approximated to any order $1/n$. –  Agustí Roig Aug 15 '10 at 15:41
@Agusti: That was my mistake in the answer (now corrected): it should have read that $p/q$ could be approximated only to order $1/b$ (where $1/b$ depends on the particular rational approximation used. Any rational number can be approximated very closely by another (this is because the rationals are dense). However, the closeness comes at the price of having a really large denominator for the approximand. (If you don't have to go really large, then you have an irrational number.) –  Akhil Mathew Aug 15 '10 at 16:14

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?

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The "main idea" you state does not seem to hold: consider the infinite sequence 1/2, 1/2 + 1/4, 1/2 + 1/4 + 1/8, 1/2 + 1/4 + 1/8 + 1/16, ... which converges to 1. –  Jason S Aug 14 '10 at 22:55
@Jason: The "main" idea fails for the series you give because the exponentials grow too slowly. –  Akhil Mathew Aug 15 '10 at 3:16
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.
If $\sum_{n \gt k} \dfrac{x_n}{n!} = 0$, can $\sum_{n \gt k+1} \dfrac{x_n}{n!} = 0$?. –  Aryabhata Dec 18 '10 at 20:56