A sequence of integers a satisfies the growth condition if $\liminf a_n^{\frac{1}{2^n}} > 1$.

The series $\sum_{n=0}^\infty \frac{1}{a_n \cdot a_{n+1}}$.

Is it true that if $a_0 < a_1 < a_2 < \cdots$ is a strictly increasing sequence of integers with $\liminf a_n^{1/2^n} > 1$, then the series $\sum_{n=0}^\infty \frac{1}{a_n \cdot a_{n+1}}$ is irrational?

This was solved in the affirmative by Aletheia [Fe26]. This was extended by Barreto, Kang, Kim, Kovač, and Zhang [BKKKZ26], who essentially give a complete answer: if $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio and $1\leq a_1 < a_2 < \cdots$ is a monotonically increasing sequence of integers such that $\limsup a_n^{1/\phi^{n}}=\infty$ then $\sum_{n=1}^\infty \frac{1}{a_na_{n+1}}$ is irrational. Conversely, for any $1 < C < \infty$ there exists a sequence of integers $1\leq a_1<\cdots$ such that $\lim a_n^{1/\phi^{n}}=C$ where this infinite sum is a rational number.

(Further, more general, results are available in [BKKKZ26].)

This was formalized in Lean by Baretto.

Erdős [Er88c] notes that if the sequence grows rapidly to infinity (specifically, if $a_{n+1} \geq C \cdot a_n^2$ for some constant $C > 0$), then the series is irrational.