Reciprocal Mersenne Subseries

Reciprocal-summable supports

For every infinite set $A$ of positive integers with $\sum _{a\in A}1/a<\mathrm{\infty }$, the series

is irrational at every integer base $b\ge 2$. This is the reciprocal-summable theorem also stated in the short paper’s Appendix A. Erdős’s 1968 paper, p. 222 proves the pairwise-coprime case at every integer base and states the removal of coprimality without printing the details. The proof below gives the stated extension by averaging; no comparison with the omitted proof is possible. The pairwise-coprime case, Theorem 6.15, retains that extra hypothesis. The same paper says that $\sum 1/n_i<\mathrm{\infty }$ could be replaced by a weaker but more complicated condition (p. 222), and suggests that for pairwise coprime supports Brun’s method could probably replace it by $\sum _{n_i<x}1/n_i=o(\log \log x)$ (p. 226). Theorem 1.1 below proves irrationality under one explicit weaker condition without coprimality; no identification with his suggestions is asserted.

We show that rationality would impose a positive lower bound and then construct smaller values. For a positive integer $N$, write $N=q_{a}a+r_a$, $0\le r_a<a$. The identity

shows that

The integer sum is finite, since $q_a=0$ for $a>N$. Every summand of ${\mathrm{\Delta }}_{b,A}(N)$ is nonnegative and an exponent $a>N$ makes it positive. Thus $X_A(b)=p/q$ would imply ${\mathrm{\Delta }}_{b,A}(N)\ge 1/q$ for every $N>0$.

Put $w_{b,d}(N)=b^{Nmodd}/(b^d-1)$. For a fixed positive modulus $Q$ and $g=(Q,d)$, the orbit of $Qmmodd$ has length $d/g$. Its geometric sum gives

To interchange this limit with the sum over the support, expand $w_{b,d}(Qm)$ as $\sum _{r\ge 1}{b}^{-r}{\mathbf{1}}_{d\mid Qm+r}$. Counting positive multiples of $d$ then gives

With $Q$ fixed, dominated convergence identifies the limiting mean of $\sum _{d\in A}{w}_{b,d}(Qm)$ with $\sum _{d\in A}{M}_{Q,b}(d)$.

Now choose $Q_t=\mathrm{lcm}(1,\dots ,t)$. For each fixed $d$, eventually $d\mid Q_t$, so $M_{Q_t,b}(d)=(b^d-1{)}^{-1}$. The bound $M_{Q_t,b}(d)\le 1/d$ permits a second dominated-convergence argument:

Consequently the limiting mean of ${\mathrm{\Delta }}_{b,A}(Q_{t}m)$ tends to zero. Given $\epsilon >0$, first choose $t$ and then a sufficiently long finite average with mean below $\epsilon$; one term of that average is below $\epsilon$. Under rationality, choosing $\epsilon <1/q$ contradicts the lower bound above. This proves the assertion. The exact all-base statement, with hypotheses $b\ge 2$, infinitude of $A$, and summability of $a\mapsto {\mathbf{1}}_A(a)/a$, has a kernel-checked Lean proof.

We increased the observation length with $Q_t$ fixed, and only then increased $t$; no independence of divisibility events was used. At base two and full support, the contribution of exponents $a>N$ alone gives ${\mathrm{\Delta }}_{2,{\mathbb{N}}_{>0}}(N)>1-2^{-N}\ge 1/2$. Full-support irrationality therefore uses a different argument; see Theorem 5.4.