Writing number as sum of reciprocals of factorial Given a real number $r>0$. Is there a way to determine whether $r$ can be written as a (possibly infinite) sum of distinct terms of the form $1/n!$?
For example, if we want to determine whether $r$ can be written as a sum of distinct terms of the form $1/3^n$, we can write $r$ in base $3$, and check whether the representation contains only digits $0$ and $1$. This question asks for a similar checking method for sum of the form $\sum 1/n!$. One necessary condition is that $r<1+1/2!+1/3!+\ldots=e-1$. But that is probably not sufficient.
 A: One necessary condition: if $r$ is such a sum, the fractional part of $n!r$ is 
$$n! r - \lfloor n!r \rfloor \le \sum_{k=n+1}^\infty \dfrac{n!}{k!} <  \dfrac{1}{n} $$
So this says something about rational approximations of $r$.
In particular, $\sin(\pi n!\; r) \to 0$ as $n \to \infty$.
Now for a randomly chosen $X \in [0,4]$, say, with uniform distribution, 
the random variables $\sin(\pi n! X)$ are identically distributed, and $\mathbb P(|\sin(\pi n! X)| < \epsilon) = O(\epsilon)$.  This implies that 
the set of $r$ with such a representation has measure $0$.
A: Let's use a conceptually similar approach to the one you proposed for base 3.
You observe that $r \leq e-1$; we can extend this a bit further:
First define $K(n) = \sum_{i=n}^\infty \frac{1}{i!}$. Note that for $n>0$, $K(n) = (e-1) - \sum_{i=1}^{n-1} \frac{1}{i!}$; and more usefully
$K(n)/n = \sum_{i=n}^\infty \frac{1}{(n)i!} > \sum_{i=n}^\infty \frac{1}{(i+1)i!} = \sum_{i=n+1}^\infty \frac{1}{i!} = K(n+1)$
We can then prove by induction that $1/(n!) > K(n+1)$ for all $n>0$:
$1/(1!) = 1 > K(2) = (e-1)-1 = 0.718...$
If $1/(n!) > K(n+1)$, then it follows that
$1/(n+1)! = 1/(n!(n+1)) > K(n+1)/(n+1) > K(n+2)$
So QED. Next, let $d_0=0$, and $r_0=r$; and for $n>0$, define:
$d_n = 1$ if $r_{n-1} \geq 1/(n!)$; else $d_n=0$
$r_n = r_{n-1} - d_n/(n!)$
Then $r$ can be expressed in the way you desire (i.e., as $r = \sum_{n=1}^\infty \frac{d_n}{n!}$) iff for all $n$, $r_n \leq K(n+1)$. 
Since $1/(n!) > K(n+1)$ for any $n$, if $r_{n-1} \geq 1/(n!)$ then $1/(n!)$ must contribute to the sequence if $r$ can possibly be represented in the way you desire; otherwise there won't be enough 'leftover' members of the summation to sum up to $r_n$.
As an example, 4/5 cannot be represented in the way you suggest: $4/5 > 1/2$, but $4/5 - 1/2 = 0.3 > K(3) = 0.2183...$
A: Not if $r \in \mathbb{R} - \mathbb{Q}$. The rationals are closed under the operation of addition.
Edit: This only applies to finite sums.
