Erdős and Graham write 'the answer is almost surely in the affirmative if $f(n)$ is assumed to be nondecreasing'. Even the case $f(n)=n$ is unknown, although Hansen [Ha75] has shown that\[\sum_n \frac{1}{\binom{2n}{n}}=\sum_n \frac{n!}{(n+1)\cdots (n+n)}=\frac{1}{3}+\frac{2\pi}{3^{5/2}}\]is transcendental.

Crmarić and Kovač [CrKo25] have shown that the answer to this question is no in a strong sense: for any $\alpha \in (0,\infty)$ there exists a function $f:\mathbb{N}\to\mathbb{N}$ such that $f(n)\to \infty$ as $n\to\infty$ and\[\sum_{n\geq 1} \frac{1}{(n+1)\cdots (n+f(n))}=\alpha.\]It is still possible that this sum is always irrational if $f$ is assumed to be non-decreasing; Crmarić and Kovač show that the set of the possible values of such a sum has Lebesgue measure zero.

View the LaTeX source

This page was last edited 28 September 2025.

1 comment on this problem

• Since in the problem variant where we assume\[ f(1)\leq f(2) \leq f(3)\leq\cdots \to\infty \]the set of the corresponding series' sums has both empty interior and zero measure, that problem variant could be thought of as being two difficulty levels above the original problem statement and one level above, say, Problem [257].

  Vjeko_Kovac — 13:09 on 09 Aug 2025

  👍 0 📝 0 🤖 0