# Sums of reciprocals involving divisor sums

Let $A\subset\mathbb{N}$ be a subset of the natural numbers, and let $\sigma(n)$ denote the sum of divisors of $n$. Recall that we have the bound $\sigma(n) = O(n\log\log n)$. Consider the sums $\sum_{a\in A} 1/a$ and $\sum_{a\in A} 1/\sigma(a)$.

Question. Is there on $A\subset\mathbb{N}$ so that the first sum diverges but the second sum converges ?

If the answer is already known, a reference would more than suffice. Thank you

Yes, this is possible. Let $k\ge 3$, say and let $P(k)$ denote the product of the first $k$ primes, and let $A_k$ denote a set of integers that are all multiples of $P(k)$ and with $$\frac{1}{k \log k} \le \sum_{a\in A_k} \frac{1}{a} \le \frac{2}{k\log k}.$$ Since the harmonic sum diverges, we can clearly choose such $A_k$, and moreover we may arrange for the sets $A_3$, $A_4$, $\ldots$ all to be disjoint (just pick $A_k$ to be the multiples of $P(k)$ in suitable disjoint intervals). Take $A$ to be the union of all the $A_k$ (with $k\ge 3$). Clearly $\sum_{a\in A} 1/a$ diverges.
Now for $a\in A_k$ we have $\sigma(a)/a \ge \sigma(P(k))/P(k) \gg \log k$ by Mertens. Therefore $$\sum_{a\in A_k} \frac{1}{\sigma(a)} \ll \frac{1}{\log k} \sum_{a\in A_k} \frac 1a \ll \frac{1}{k (\log k)^2}.$$ Therefore $$\sum_{a\in A} \frac{1}{\sigma(a)}$$ converges.
• Nice one (+1): picking $A_k$ as multiples of $P(k)$ and $1$ modulo the $k+1$-th prime, could be also a way too. Sep 1, 2015 at 22:46