On the irrationality of $\sum_{n=1}^{\infty} b^{-a_n} $ In my answer to
Hardy and Wright irrational sums
I showed that
if
$a_n$
is a positive increasing
sequence of integers and
$\dfrac{n}{a_n} 
\to 0
$
and
$b \ge 2$
is an integer
then
$S
=\sum_{n=1}^{\infty} b^{-a_n}
$
is irrational.
However,
the condition
$\dfrac{n}{a_n} 
\to 0
$
is sufficient,
not necessarily necessary.
Since
$a_n \ge n$
by assumption,
$\dfrac{n}{a_n} 
\le 1
$.
My question:
Is there a sequence of $a_n$
such that
$\dfrac{n}{a_n} 
\to 1
$
and
$S
=\sum_{n=1}^{\infty} b^{-a_n}
$
is irrational?
 A: Let $c_n$ be any increasing sequence of positive integers such that $\frac{n}{c_n} \to 0$. Let $a_n$ be the increasing sequence of positive integers such that $\mathbb{Z}_{+}$ is a disjoint union of $\{ a_n : n \in \mathbb{Z}_{+} \}$ and $\{ c_n : n \in \mathbb{Z}_{+} \}$. It is easy to see $\frac{n}{a_n} \to 1$.
Notice
$$\sum_{n=1}^\infty b^{-a_n} + \sum_{n=1}^\infty b^{-c_n} = \sum_{n=1}^\infty b^{-n} = \frac{1}{b-1} \in \mathbb{Q}$$
and you have shown $\sum_{n=1}^\infty b^{-c_n}$ is irrational, this implies $\sum_{n=1}^\infty b^{-a_n}$ is also irrational.
A: Yes. There are uncountably many density $0$ subsets of $\mathbb{N}$, each gives rise to a sequence $(a_n)_n$ such that $\frac{n}{a_n} \to 1$, and each gives rise to a different value of $S$, but there are only countably many rationals.
A: My answer is 
based on the result
in my proof that
the number is irrational
if and only if
$a_{n+1}-a_n$
is unbounded.
Let
$v_n$ be a fast-growing sequence
such as $v_n = 2^n$
and define $a_n$ by
$a_{n+1}=a_n+\ln(n)$
if $a_n = v_k$
for some $k$,
and
$a_{n+1}=a_n+1$
otherwise.
Then
$a_{n+1}-a_n$
is unbounded,
but gets large not often
and very slowly,
so it is easy to show that
$\dfrac{n}{a_n}
\to 1
$.
