irrationality of the series $\sum \frac{1}{2^{n!}}$ How can I prove that the series
$\sum_{n=0}^{\infty} \frac{1}{2^{n!}}$ converges to an irrational number?
 A: Its binary expansion never repeats.
A: Let $(a_n)$ an increasing sequence of positive integers.
Suppose that $\sum_n \frac{1}{2^{a_n}}=\frac{p}{q}$. Then, for any $m$, we have $$\sum_n \frac{q2^{a_m}}{2^{a_n}}=p2^{a_m}\in\mathbb{Z}$$
Since $\frac{q2^{a_m}}{2^{a_n}}\in\mathbb{Z}$ for $n\leq m$, we conclude  $$\sum_{n>m}\frac{q2^{a_m}}{2^{a_n}}\in\mathbb{Z}$$
In particular
$$\sum_{n>m}\frac{q}{2^{a_n-a_m}}\geq 1,$$ This would imply that
$$\sum_{n>m}\frac{1}{2^{a_n-a_m}}\geq 1/q\hbox { for all } m$$.
Now, when $a_n=n!$ what happens is that this sequence increases too fast, in particular note that in this case $$\sum_{n>m}\frac{1}{2^{n!-m!}}\leq \sum_{n>m}\frac{1}{2^{n}}=\frac{1}{2^{m}}$$
So, for big enough $m$ you will have 
$$\sum_{n>m}\frac{1}{2^{n!-m!}}<\frac{1}{2^{m}}<1/q$$.
This ("contradiction") would imply that $\sum_n\frac{1}{2^{n!}}$ can't be a rational number.
Remarks:
This kind of proof works for any kind of series $\sum \frac{1}{b_n}$ where $b_n$ are integers increasing fast with $b_i|b_{i+1}$.  For instance, you may prove that $e$ is irrational in this way. 
The number in your example can actually be proven to be trascendental (related to Liouville number).
A: As mentioned by @Athur, the problem has a simple answer by thinking of the infinite acyclic binary expansion of $\sum_n\frac{1}{2^{n!}}$. I would like however to comment on  a slightly different problem which contains this OP as a particular case:

Suppose $n_k$ is a monotone increasing sequence of integers such that $\limsup_k(n_{k+1}-n_k)=\infty$, and $\varepsilon:\mathbb{N}\rightarrow\{-1,1\}$. Let $x=\sum^\infty_{k=1}\frac{\varepsilon_k}{2^{n_k}}$. Then $x$ is irrational.

Here we appeal to the following well known result for which I present a short proof at the end:

Theorem: For any $x\in\mathbb{R}$, $x\in\mathbb{R}\setminus\mathbb{Q}$ iff there exists a sequence of integers $p_k\in\mathbb{Z}$, $q_k\in\mathbb{N}$ such that $|q_k x - p_k|>0$ and $\lim_{k\rightarrow\infty}|q_k x - p_k|=0$.

Consider the partial sum $s_k=\sum^k_{j=1}\frac{\varepsilon_j}{2^{j}}$. Then
$$ s_k=\frac{\varepsilon_12^{n_k-n_1}+\ldots+\varepsilon_{k-1}2^{n_k-n_{k-1}}+\varepsilon_k}{2^{n_k}}=\frac{p_k}{2^{n_k}}$$
is an irreducible fraction, and
$$\Big|x-\frac{p_k}{2^{n_k}}\Big|=|x-s_k|=\sum_{j>k}\frac{1}{2^{n_j}}\leq \sum^\infty_{j={n_{k+1}}}\frac{1}{2^j}=\frac{2}{2^{n_{k+1}}}=2\frac{2^{-n_k}}{2^{n_{k+1}-n_k}}$$
The assumption $\limsup_k(n_{k+1}-n_k)=\infty$ means that along some subsequence $k'\nearrow\infty$
$$\Big|x-\frac{p_{k'}}{2^{n_{k'}}}\Big|=o\Big(\frac{1}{2^{n_{k'}}}\Big)$$
From this, we conclude that $x$ is irrational.
For the OP, consider $\varepsilon_k\equiv1$ and notice that $n_k=k!$ satisfies the conditions of the Theorem.

Proof of Theorem:
Without loss of generality, assume that $x\in[0,1)$ (otherwise that the fractional part $\{x\}=x-\lfloor x\rfloor$.
Claim: for any $N>0$, there exists $p,q\in\mathbb{Z}$, $0<q\leq N$ such that $|qx-p|<\frac{1}{N}$. Split $[0,1)$ in $N$ pieces of length $1/N$. Consider the numbers $\delta_j=\{jx\}$, $j=0,\ldots, N$. There must be an subinterval, that contains at least two elements say, $\delta_j$, $\delta_k$ with $0\leq j<k\leq N$. Then
$$|\delta_k-\delta_j|=|(k-j)x-(\lfloor kx\rfloor-\lfloor jx\rfloor)|<\frac{1}{N}$$
taking $q_N=(k-j)$ and $p_N=(\lfloor kx\rfloor-\lfloor jx\rfloor)$ we obtain $0<q_N\leq N$ and
$$ \big|q_N x-p_N|<\frac{1}{N}$$
Necessity: Suppose $x$ is irrational. By the claim, for any $k\in\mathbb{N}$, there are $p_k,q_k\in\mathbb{Z}$ and such that $0<q_k\leq k$, and  $0<|p_k x - q_k|<\frac1k$, where  positivity follows from the fact that $x$ is assumed to be irrational.  Since $\lim_k|p_kx-q_k|=0$, it follows that there are infinitely many distinct $(p_k,q_k)$. Furthermore, if $\alpha_k=\operatorname{g.c.d}(p_k,q_k)$, then
$$0<|q'_k x- p'_k|<\frac{1}{\alpha_k k}\leq \frac1k$$
and so, the pairs $(p_k,q_k)$ can be taken so that $\tfrac{p_k}{q_k}$ are irreducible, and $q_k\nearrow\infty$.
Sufficiency: Suppose $x\in\mathbb{R}$ is such that there are $p_k\in\mathbb{Z}$ and $q_k\in\mathbb{N}$ satisfying  $|q_k x-p_k|>0$ and $\lim_k|q_k x - p_k|=0$. If $x=\frac{p}{q}\in\mathbb{Q}$, then
$q|q_k x-p_k|=|q_kp-p_kq|\in\mathbb{N}$,  and $\lim_k|q_k p - p_kq|=0$. Consequently, for all $k$ large enough $p_kq-p_kq=0$. This is in contradiction to the assumption that all $|q_k x-p_k|=\frac{1}{q}|q_k p - p_kq|>0$ for all $k$.
