Let $K$ be a number field with norm $N=N_{K/\mathbb{Q}}$, $\mathcal{P}$ its set of prime ideals, and $A\subset\mathcal{P}$. We say that the Dirichlet density is $$\delta^D(A):=\lim\limits_{s\to1^+}\frac{\sum\limits_{p\in A}N(p)^{-s}}{-\log(s-1)}$$ and the polar density $\delta^P(A)$ is $\frac{m}{n}$, where $A$'s Dedekind zeta function $\zeta_A(s):=\prod\limits_{p\in A}(1-N(p)^{-s})^{-1}$ has its $n$th power with a pole of order $m$. For each of these notions of density, what is an example of a set of primes on which the densities are not defined?
I know that such sets must exist via the theorem "if $\delta^P(A)$ exists then $\delta^D(A)$ exists and they are equal" so by implication the hypothesis must not always hold. For an instance where $\delta^P$ does not exist, it would be especially intriguing if $\delta^D$ did exist.
One example along these lines is for natural density, $$\delta^N(A):=\lim\limits_{x\to\infty}\frac{\#\{p\in A:N(p)<x\}}{\#\{p\in\mathcal{P}:N(p)<x\}}:$$ enumerate $\mathcal{P}$ by increasing norm and going down the list, include enough primes to bring the quotient to above $\frac{3}{4}$, then ignore enough primes to being it below $\frac{1}{4}$, then continue alternating indefinitely, so that the limit does not actually exist. Could something like this work for $\delta^D$?
It works the same way for $A$ a set of integer primes, and yes the same idea works for the weaker density $\lim\limits_{s\to1^+}\frac{\sum\limits_{p\in A}N(p)^{-s}}{-\log(s-1)}$.
This is because $\frac{\sum\limits_{p\in A}N(p)^{-s}}{-\log(s-1)}$ can be $\epsilon$ approximated by just knowing the elements of $A$ whose norm is $\in [l_{s,\epsilon} , L_{s,\epsilon}]$ with $$\lim_{s\to 1}l_{s,\epsilon} =\infty,\quad \lim_{s\to 1} L_{s,\epsilon}=\infty$$
Fix $\epsilon=10^{-2}$, choose $s_k\to 1$ such that $l_{s_k,\epsilon} >L_{s_{k-1},\epsilon}$, you will be able to construct $A$ such that $$\frac{\sum\limits_{p\in A}N(p)^{-s_k}}{-\log(s_k-1)}\in \frac{1}2+\frac{(-1)^k}4+[-\epsilon,\epsilon] $$