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Source. Theorem 3.1, printed p. 87; proof pp. 87--88. Read on the page images.

Statement

Write $p_n$ for the $n$-th prime, and let the positive integers $a_n$ form a monotonic sequence with

$$\lim_{n\to\infty}\frac{p_n}{a_n^2}=0\qquad\text{and}\qquad \liminf_{n\to\infty}\frac{a_n}{p_n}=0 .$$

Then the sum

$$\sum_{n=1}^{\infty}\frac{p_n}{a_1\cdots a_n}\qquad(3.2)$$

is irrational.

Proof structure (pp. 87--88)

The series satisfies the hypotheses of Theorem 2.1, so rationality gives $B$ and integers $c_n$ with (3.3) $Bp_n=c_na_n-c_{n+1}$ for large $n$. If $c_n=c_{n+1}$ held at some large $n$, then $c_n\mid B$ and $a_n>p_n$; as the $c_n$ are unbounded, a later index $m$ with $c_m\le c_n<c_{m+1}$ would then give (3.4) $p_{m+1}>p_m+a_m/(2B)>(1+1/(2B))p_m$, impossible for large $m$; so $c_n\ne c_{n+1}$ for large $n$. Now let $n$ run over $(N,2N)$ for a large $N$. A rise $c_{n+1}>c_n$ gives, as in (3.4), $p_{n+1}>p_n+a_n/(2B)>p_n+\sqrt{p_n}$; these gaps add up to at most $p_{2N}-p_N$, so there are fewer than $(p_{2N}-p_N)/\sqrt{p_N}<N^{1/2+\varepsilon}$ rises. All remaining $n$ are falls $c_{n+1}<c_n$, and each fall gives (3.5) $a_{n+1}>a_n+(a_n-1)/c_{n+1}>a_n+1$. Falls fill most of the range, so $a_{2N}>N/2$; hence $a_n>n/4$ and $c_n<p_n/a_n+1<\sqrt n/4$ once $n$ is large, and (3.5) sharpens to (3.6) $a_{n+1}>a_n+\sqrt n$ at each large fall. Then $a_{2N}>N^{3/2}/2$, against $\liminf a_n/p_n=0$.

The prime input is mild but includes a bound on individual gaps: (3.4) is ruled out for large $m$ only because $p_{m+1}/p_m\to1$, and the rest of the argument, including the count of large gaps in $(N,2N)$, uses $p_n\ll n\log n$. Both follow from $p_n\sim n\log n$; no gap bound as strong as $p_{n+1}-p_n=o(n)$ is used.

Specialization to the factorial series

With $a_n=n$: the sequence is monotone, $p_n/n^2\to0$ and $\liminf n/p_n=0$ hold since $p_n\sim n\log n$; hence $\sum p_n/n!$ is irrational, the case $k=1$ of Erdős 1958, here without any prime-gap hypothesis. The condition $p_n=o(a_n^2)$ excludes bounded $a_n$, in particular $a_n=2$; the theorem says nothing about $\sum p_n/2^n$.

Later strengthening

Hančl–Tijdeman 2004, Theorem 5.1 drops $\liminf a_n/p_n=0$: for monotone positive integers $a_n$ with $p_n=o(a_n^2)$, the series is rational if and only if $p_n/(a_n-1)$ is eventually constant; their Theorem 6.1 weakens $p_n=o(a_n^2)$ to $a_n/\log n\to\infty$.

Bears on. #251 (context: the monotone relatives of the problem's series, and a reproof of the $k=1$ theorem cited on the problem page).