TITLE: Irrationality proof technique: no factorial in the denominator
QUESTION [19 upvotes]: Jonathan Sondow elegantly proves the irrationality of e in his aptly titled A Geometric Proof that e Is Irrational and a New Measure of Its Irrationality (The American Mathematical Monthly, Vol. 113, No. 7 (Aug. - Sep., 2006), pp. 637, http://www.jstor.org/stable/27642006).
In his argument, he constructs a sequence of nested intervals $I_n$ for every $n \geq 1$, each of the form $[k/n!, (k+1)/n!]$, such that $\bigcap I_n = \{e\}$, with $e$ lying strictly between the endpoints of each $I_n.$ From this, we conclude that $e$ cannot be written as a fraction with denominator $n!$ for any $n \geq 1.$
Fact: Every rational number $p/q$ can be written as a fraction with a factorial in its denominator: $p/q = p(q-1)!/q!$. 
Thus, we conclude that $e$ is irrational.
The reason this proof technique works so well with $e$ is, of course, related to the Maclaurin series for the exponential function, $e^x.$
That any rational number can be written in lowest terms is employed in other irrationality proofs (e.g., the classic proof for that of $\sqrt{2}$) but I had not seen the above fact drawn upon before reading this particular paper.
My question is: are there other examples of real numbers (which are not related to $e$ in some trivial way) whose irrationality can be proved using the Fact above?

REPLY [5 votes]: For any interval $I_n$ of the form you define, there are $n+1$ intervals that could be $I_{n+1}$ nested in it. Choose one of those intervals, then repeat the process. As long as you choose an interval other than the first infinitely many times and an interval other than the last infinitely many times, you get an irrational number by this proof.
Or in other words, we have:
$$\sum_{m=n+1}^\infty \frac{m-1}{m!} = \sum_{m=n+1}^\infty \left(\frac{1}{(m-1)!} - \frac{1}{m!}\right) = \frac{1}{n!}$$
so in user22202's answer, for any sequence of integers $a_m$ such that $0 \leq a_m \leq m-1$, and $a_m>0$ infinitely often and $a_m < m-1$ infinitely often, we have
$$ \sum_{m=0}^\infty \frac{a_m} {m!}$$
is irrational.