Dirichlet vs. logarithmic density 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?

 A: 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.
