TITLE: Dirichlet vs. logarithmic density
QUESTION [8 upvotes]: The Dirichlet density of A relative to B is 
$$
\lim_{s\to 1^+}\frac{\sum\limits_{n\in A}n^{-s}}{\sum\limits_{n\in B}n^{-s}}
$$
and the logarithmic density of A relative to B is
$$
\delta(A) = \lim_{n \to \infty} \frac{\sum_{a \in A,\, a \le n} \frac1a}{\sum_{b\in B,b\le n}\frac 1b}
$$
with the respective lower and upper densities defined with $\liminf$ and $\limsup$ respectively. ($A\subseteq B\subseteq\mathbb{N}$ of course.)
I have read in one source that the Dirichlet density exists if and only if the logarithmic density exists (in which case they are equal). In another source I read that they are the same. I wanted to check; to wit:

Is the lower Dirichlet density always equal to the lower logarithmic density?
Is the upper Dirichlet density always equal to the upper logarithmic density?
Does the Dirichlet density always exist if and only if the lower logarithmic density exists?
If both exist, are they equal?
Do the above hold in the special case $B=\mathbb{N}$?
Sometimes terminology is not standardized; are these definitions standard?

REPLY [3 votes]: Let
$$ \alpha_{n} = \sum_{\substack{ a \leq n \\ a \in A}} \frac{1}{a}
\quad \text{and} \quad
\beta_{n} = \sum_{\substack{ b \leq n \\ b \in B}} \frac{1}{b}. $$
The following holds:
$$ \varliminf_{n\to\infty} \frac{\alpha_{n}}{\beta_{n}}
\leq \varliminf_{s\to 0^{+}} \frac{\sum_{n\in A} n^{-1-s}}{\sum_{n\in B} n^{-1-s}}
\leq \varlimsup_{s\to 0^{+}} \frac{\sum_{n\in A} n^{-1-s}}{\sum_{n\in B} n^{-1-s}}
\leq \varlimsup_{n\to\infty} \frac{\alpha_{n}}{\beta_{n}} \tag{1} $$
It is clear that this inequality holds if $\beta_{n}$ converges. (In this case, the above inequality reduces to equality.) So we may assume that $\beta_{n} \to \infty$. Then note that we have
$$ \sum_{n\in A} \frac{1}{n^{1+s}} = \sum_{n=1}^{\infty} \alpha_{n} \left( \frac{1}{n^{s}} - \frac{1}{(n+1)^{s}} \right) $$
and likewise for $\beta_{n}$. Thus if we put $\bar{\rho} = \varlimsup \alpha_{n} / \beta_{n}$, for $\epsilon > 0$ we can find $N$ such that $\alpha_{n} < (\bar{\rho} + \epsilon) \beta_{n}$ for $n > N$. Thus
\begin{align*}
\sum_{n\in A} \frac{1}{n^{1+s}}
&\leq \sum_{n\leq N} \alpha_{n} \left( \frac{1}{n^{s}} - \frac{1}{(n+1)^{s}} \right) + \sum_{n>N} (\bar{\rho} + \epsilon) \beta_{n} \left( \frac{1}{n^{s}} - \frac{1}{(n+1)^{s}} \right) \\
&= \sum_{n\leq N} (\alpha_{n} - (\bar{\rho} + \epsilon) \beta_{n}) \left( \frac{1}{n^{s}} - \frac{1}{(n+1)^{s}} \right) + (\bar{\rho} + \epsilon) \sum_{n\in B} \frac{1}{n^{1+s}} \tag{2}
\end{align*}
Dividing both sides of $(2)$ by $ \sum_{n\in B} n^{-1-s} $ and taking $s \to 0^{+}$, we have
$$ \varlimsup_{s\to 0^{+}} \frac{\sum_{n\in A} n^{-1-s}}{\sum_{n\in B} n^{-1-s}}
\leq \bar{\rho} + \epsilon. $$
Here, we exploited the fact that
$$ \lim_{s \to 0^{+}} \sum_{n\in B} \frac{1}{n^{1+s}} = \sum_{n\in B} \frac{1}{n} = \lim_{n\to\infty} \beta_{n} = \infty $$
so that the $\sum_{n \leq N}$ part of the sum $(2)$, divided by $ \sum_{n\in B} n^{-1-s} $, vanishes as $s \to 0^{+}$. Since $\epsilon$ is arbitrary, we obtain the rightmost part of the inequality $(1)$. Similar consideration proves another side, completing the proof.
Kunnysan showed that $(1)$ need not reduce to an equality, hence this is not far from the sharpest result.