Consider the following ways to average a multiplicative arithmetic function $f$ over $\mathbb{N}$:
- Arithmetic: $\displaystyle\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N f(n)$
- Factorized: $\displaystyle\prod_p \left(1-\frac{1}{p}\right)\sum_{v=0}^\infty \frac{f(p^v)}{p^v}$
- Regularized: $\displaystyle\lim_{s\to1^+}\frac{1}{\zeta(s)}\sum_{n=1}^\infty \frac{f(n)}{n^s}$
- Harmonic: $\displaystyle\frac{1}{H_N}\sum_{n=1}^N \frac{f(n)}{n}$
Dropping the multiplicativity condition, if $f$ is an indicator function of a set $S\subseteq\mathbb{N}$ then the arithmetic and regularized averages are called the arithmetic and Dirichlet densities respectively. I believe if the arithmetic density exists then the Dirichlet one does as well and then they're equal, but not conversely.
Question A. Is it possible for two of these averages to exist but not be equal?
Question B. For which pairs of averages does the existence of one imply the existence of the other (possibly under an extra assumption on the regularity or growth of $f$)?
The hope is that there is literature on this already.
The factorized average is my own invention, based on the heuristic that a "random" natural number $n$ is an infinite product of powers $p^{v_p}$, where each exponent $v_p$ is random with a pdf that says the probability that $v_p=v$ is the (arithmetic) density of the set of numbers $n$ with $v_p(n)=v$, which is $(1-1/p)p^{-v}$. The probability that infinitely many $v_p$s are nonzero is $0$, and if $f$ is multiplicative we use the heuristic $\mathbb{E}(f(n))=\prod_p \mathbb{E}(f(p^v))$ based on divisibility by different primes being independent.
According to this heuristic, the probability $n$ is a specific natural is $0$ since $\prod_p(1-1/p)=1/\zeta(1)=0$, but the ratio between the probabilities for getting specific naturals $m$ and $n$ ought to be like the ratio $(1/m)/(1/n)$, which motivates using limiting cases of the zeta and Zipf distributions for the regularized and harmonic averages.