Problem #270: Let $f(n)\to \infty$ as $n\to \infty$

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Progress

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.