Irrationality of certain lacunary series This question is motivated primarily by the following postings:

*

*A series of rational number converges to an irrational number

*The irrationality of rapidly converging series

*Is $\sum\limits_{n=1}^\infty\frac1{a_n}$ irrational? 
among others,  which concern the irrationality of series of the form $\sum^\infty_{k=1}\frac{1}{n_k}$ where  $n_k$ is is a very lacunary monotone increasing sequence of of positive integers.

My precise question is as follows. Suppose $(\varepsilon_k:k\in\mathbb{N})$ is a sequence taking values in $\{-1,1\}$, and $n_k$ is lacunary monotone increasing sequence of integers. Is
$$x:=\sum^\infty_{k=1}\frac{\varepsilon_k}{n_k}$$
irrational? The answer in general is no unless the gaps between the terms $n_k$ and $n_{k+1}$ satisfy some additional conditions. Probably, the easiest instance of this problem is when
$$\begin{align}
n_k:=2^{m_k},\quad\text{where}\quad \limsup_k(m_{k+1}-m_k)=\infty\tag{0}\label{zero}\end{align}$$
in which case $x$ is irrational. Other cases of interest are when $n_k$ is such that either
$$\begin{align} \liminf_k\frac{n_1\cdot\ldots\cdot n_k}{n_{k+1}}=0,\quad\text{and}\quad n_{k+1}\geq C n_k\tag{1}\label{one}\end{align}$$
and
$$ \begin{align}\limsup_k\sqrt[2^k]{n_k}=\infty,\quad\text{and}\quad n_{k+1}\geq C n_k\tag{2}\label{two}\end{align}$$
for some content $C>1$. That $x$ is irrational in  this cases got registered in notes  I took long time ago during a talk about lacunary random trigonometric series (I am sure it is a well known fact for analytic number theorists) and just resurfaced a few days ago while organizing papers.

As a general method, It is enough to show that there are infinitely pairs of integers $(p_k,q_k)$, with $q_k>0$, such that $0<|q_k x -p_k|$ and $\lim_{k\rightarrow\infty}|q_k x- p_k|=0$. (This is known to be a necessary and sufficient condition for irrationality).
For example, under assumption \eqref{one}, if
$s_k=\sum^n_{j=1}\frac{\varepsilon_k}{n_k}=\frac{p_k}{n_1\cdot\ldots\cdot n_k}=\frac{p_k}{q_k}$, then
$$|q_k x - p_k|\leq q_k\sum^\infty_{j=k+1}\frac{1}{n_j}\leq \frac{n_1\cdot\ldots\cdot n_k}{n_{k+1}}\sum^\infty_{n=1}C^{-n}\xrightarrow{k\rightarrow\infty}0$$
along a subsequence $k$. Thus, in this case $x$ is irrational.

Edit: Originally I had typed condition \eqref{two} as
$$ \begin{align}\limsup_k\sqrt[k]{n_k}=\infty,\quad\text{and}\quad n_{k+1}\geq C n_k\tag{2'}\label{twop}\end{align}$$
As it was pointed out by Erick Wong, that condition it is not enough on its own. On reviewing my notes again, I realized that the right root was $2^k$, not $k$. Incidentally, in Eric Wong's example the lacunary sequence  has the $2^k$ growth threshold.
Still with this correction I am no able to put an end to the case \eqref{two}. Any hints and/or solutions are welcome. If someone even can provide conclusion on whether the types of series described above are transcendental or not, even better!
 A: I think this is false as written (perhaps a typo in your notes or a missing condition like divisibility): let $n_1 = 2$ and $n_{i+1} = f(n_{i})$ where $f(x) = x^2 - x + 1$, so $n_2 = 3, n_3 = 7, n_4 = 43$, etc.
We easily calculate that $\frac1n = \frac{1}{n-1} -  \frac{1}{f(n)-1}$ and hence telescopically,
$$\sum_{i=1}^k \frac{1}{n_i} = \frac{1}{2-1} - \frac{1}{n_{k+1} - 1},$$
so the series rapidly converges to the very rational number $1$.  Meanwhile, the denominators grow superexponentially, essentially squaring each time.  Using the inequality $f(n) - \tfrac12 > (n - \tfrac12)^2$, we see that $n_k > C^{2^k}$ for some $C>1$.  This is far faster growth than needed to show that $\sqrt[k]{n_k} \to \infty$ and $n_{k+1} / n_k\to \infty$.
(This is the infinite version of the classic method of writing $1$ as an Egyptian fraction with arbitrarily many terms.)
I believe we can recover the irrationality of $x$ by adding a quite strong condition like $n_k \mid n_{k+1}$, which mimics the conditions of the proof of irrationality of $e$ by making $n_k$ itself a viable denominator for approximation.
EDIT: Now that Oliver has identified the correct statement of the question, I think the intention is for this to follow from the condition (1).  In any sequence of reals $\{a_n\}$ which is not bounded from above, there must exist infinitely many indices $k$ such that $a_{k+1} > \max \{ a_i : 1 \le i \le k\} $, and that subsequence of terms necessarily diverges to $+\infty$.  Suppose $k$ is one such index for the sequence $\{ n_r^{1/2^r} \}$.  Then for some $c:=c_k>1$ we have $n_{k+1} = c^{2^{k+1}}$, but $n_i < c^{2^i}$ for all $i \le k$.  This means
$$ \frac{n_1 n_2 \cdots n_k }{n_{k+1}} < \frac {c^2 c^4\cdots c^{2^k}}{c^{2^{k+1}}} = \frac1{c^2}.$$
And by construction, $c_k \to \infty$ along the selected subsequence, so we recover the property (1).
EDIT 2: By request, I’ll try to shed some light on my thinking that led to this counterexample.  It is an old chestnut to show that every rational number can be written as a sum of distinct unit fractions (aka Egyptian fractions).  The standard proof of this is to use a greedy algorithm: keep subtracting the largest available $1/n$ and show by an easy computation that the numerator is (eventually) decreasing.
But there’s another proof that works by starting with $a/b = 1/b + \cdots + 1/b$ and then repeatedly applying the substitution $1/n \mapsto 1/(n+1) + 1/(n^2+n)$ to resolve duplicates.  It takes a lot more effort to show that there is a finite sequence of substitutions that eliminates all duplicates, and one of the key ingredients is the fact that the terms grow so quickly.  I didn’t think of this construction when I first read your question, but that was the a-ha moment that led to the solution.
Years ago, I once marked for a course in math history, and one assignment was about Egyptian fractions (a topic that, while I haven’t studied deeply myself, runs in my mathematical blood).  Some students attempted to argue this by a different substitution: $1/n \mapsto 1/(2n) + 1/(3n) + 1/(6n)$.  It was a satisfying exercise to show that this strategy is provably destined to fail: for sufficiently large $a$, there is no way to repeatedly apply this substitution to eliminate all $a$ duplicates (some $a - O(1)$ terms will always collide).  In some sense this sequence doesn’t spread out fast enough to avoid clumping together.
