On the irrationality of $\sum_{n=1}^{\infty} b^{-a_n} $

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $.