The summand $\frac{1}{n^{1+it}}$, for a natural number $n$ and $t\in\mathbb{R}$.

Let $1<a_1<\cdots$ be a sequence of integers such that $\sum\frac{1}{a_i}<\infty$. Is it true that, for every $t\in \mathbb{R}$, $$1+\sum_{k}\frac{1}{a_k^{1+it}}\neq 0?$$

Yip [Yi25] has proved that this is not always true - in fact, for any real $t\neq 0$, there exists a sequence of integers $1<a_1<\cdots$ such that $\sum \frac{1}{a_i}<\infty$ and $1+\sum_{k}\frac{1}{a_k^{1+it}}=0$.

This was formalized in Lean by Wu using Aristotle.

Yip [Yi25] has proved that this is not always true - in fact, for any real $t\neq 0$, there exists a sequence of integers $1<a_1<\cdots$ such that $\sum \frac{1}{a_i}<\infty$ and $1+\sum_{k}\frac{1}{a_k^{1+it}}=0$.

It remains open whether this is true for every finite sequence of integers.

A question of Erdős and Ingham [ErIn64]. The simplest case they could not decide this question for was the finite sequence ${2,3,5}$.

Their interest in this problem arose from their proof that the statement that there are no such zeros is equivalent to the fact that, for any non-decreasing $f:\mathbb{R}\to \mathbb{R}_{\geq 0}$ which is bounded on every bounded interval and is $=0$ for $x<1$, the relationship $$f(x)+\sum_k f(x/a_k)=\left(1+\sum_k \frac{1}{a_k}+o(1)\right)x$$ implies $f(x)=(1+o(1))x$.