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Claim. Steve Fan's post of 11 October 2025 in the discussion thread of Problem 269 proves that the sum is irrational when $P$ is an infinite set of primes. With $q_1<q_2<\cdots$ the powers of the primes in $P$ and $x_i$ the lcm of $q_1,\ldots,q_i$, it writes the sum as $\sum_i m_i/x_i$, where $m_i$ counts the $a_n$ in $[q_i,q_{i+1})$. It shows $m_i=o(q_i)$, by Brun's sieve when $\sum_{p\notin P}1/p$ diverges and by the prime number theorem with the classical error term when it converges. If the sum were a rational $a/b$, then for large $N$ with $q_{N+1}\in P$ the integer $bx_N\sum_{i\ge N+1}m_i/x_i$ would be $o(1)$, which the post takes as the contradiction.

Submission note. Posted to the erdosproblems.com forum by Steve Fan (account Steve_Fan) on 11 October 2025. The post was edited after the curator's reply of the same day; the text below is the post as it read on 2026-10-18:

Here is a proof for the case where $P=\{p_1<p_2<\cdots\}$ is infinite. Denote the series in consideration by $S$. Let $\{q_i\}_{i\ge1}$ be the ascending sequence of the powers of primes in $P$, and let $x_i:=[q_1,...,q_i]$ for $i\ge1$. For each prime power $q=p^k$, define $\alpha(q):=p$. Then $x_{i+1}=\alpha(q_{i+1})x_i$ and $q_{i+1}\le p_1q_{i}$ for all $i\ge1$. Observe that

$$> S=\sum_{i\ge1}\frac{m_i}{x_i}, >$$

where $m_i$ counts the number of $a_n$'s in $[q_i,q_{i+1})$.

It is not hard to see that $m_i=o(q_i)$. Indeed, when $\sum_{p\not\in P}\frac{1}{p}=\infty$, Brun's sieve yields

$$> m_i\ll(q_{i+1}-q_i)\prod_{\substack{p\le q_{i+1}-q_i\\p\notin P}}\left(1-\frac{1}{p}\right). >$$

So for sufficiently large $i$, either $q_{i+1}-q_i=o(q_i)$ or the product above is $o(1)$. In both cases we have $m_i=o(q_i)$. When $\sum_{p\not\in P}\frac{1}{p}<\infty$, the Prime Number Theorem with the classical error term gives

$$> \sum_{\substack{p\le x\\p\in P}}\frac{1}{p}=\log\log x+C+O\left(e^{-c_0\sqrt{\log x}}\right) >$$

with some absolute constants $C$ and $c_0>0$, which implies particularly that given any $\epsilon>0$, every interval $(x,(1+\epsilon)x]$ with $x$ sufficiently large contains a prime in $P$, so that $m_i\le q_{i+1}-q_i\le\epsilon q_i$.

Assume now that $S=\frac{a}{b}$, where $a,b\in\mathbb{N}$. Suppose that $q_{N+1}\in P$ . Then

$$> bx_N\sum_{i\ge N+1}\frac{m_i}{x_i}=ax_N-bx_N\sum_{i\le N}\frac{m_i}{x_i}\in\mathbb{N}. >$$

Since $q_i\le p_1^{i-N-1}q_{N+1}=p_1^{i-N-1}\alpha(q_{N+1})$, we have

$$> bx_N\sum_{i\ge N+1}\frac{m_i}{x_i} =o\left(\sum_{i\ge N+1}\frac{q_i}{\alpha(q_{N+1})\cdots \alpha(q_{i})}\right)=o\left(\sum_{i\ge N+1}\frac{p_1^{i-N-1}}{\alpha(q_{N+2})\cdots \alpha(q_{i})}\right) >$$

for sufficiently large $N$. Note that

$$> \sum_{i\ge N+1}\frac{p_1^{i-N-1}}{\alpha(q_{N+2})\cdots \alpha(q_{i})}\le\sum_{i\ge N+1}\left(\frac{p_1}{p_2}\right)^{n_i}=\sum_{k\ge0}\left(\frac{p_1}{p_2}\right)^{k}\sum_{\substack{i\ge N+1\\n_i=k}}1, >$$

where $n_i:=\#\{j\in[N+2,i]\colon\alpha(q_j)>p_1\}$. Since each interval $(p_1^{r},p_1^{r+l}]$ with $l=\lceil\frac{\log p_2}{\log p_1}\rceil$ contains a power of $p_2$, we have

$$> \sum_{\substack{i\ge N+1\\n_i=k}}1\le l+1, >$$

so that

$$> \sum_{k\ge0}\left(\frac{p_1}{p_2}\right)^{k}\sum_{\substack{i\ge N+1\\n_i=k}}1\le(l+1)\sum_{k\ge0}\left(\frac{p_1}{p_2}\right)^{k}\ll1. >$$

So

$$> bx_N\sum_{i\ge N+1}\frac{m_i}{x_i} =o(1) >$$

for sufficiently large $N$, a contradiction.

Why it is rejected. It answers a variant, not the Statement. Problem 269 asks about a finite set of at least two primes. erdosproblems.com records the infinite case apart from the question, as always irrational, and reports that Erdős calls it a simple exercise in [Er88c]. A proof for infinite $P$ says nothing about any finite $P$, so its mathematics answers the variant and does not settle this problem; the value proved is its answer to that variant.

The thread. The same day Thomas Bloom, the erdosproblems.com curator, replied to the post's first version that $x_{n+1}\ge2x_n$ does not hold for all $n$, failing for example when $P$ is the set of all primes, and that the proof could be amended to show the convergence of the final tail series directly. Fan replied that he would fix it, and the post was edited; Bloom's reply now carries a note that it answered the earlier version.

Standing. Rejected, as answering a variant. The erdosproblems.com page attributes the infinite case to Erdős and does not mention the post. Bloom's reply corrected the first version and is not a review of the edited proof, and this corpus has not checked it.

Depends on. No page of this wiki.