Dirichlet density alone does not give a counting asymptotic

Write $s=1+ϵ>1$. For a decreasing function $t^{-s}$, the discrepancy between its sum and integral on the kth block is at most ${10}^{-ks}$. Summing these discrepancies is O(1) uniformly as $ϵ\downarrow 0$. The integrals form a geometric series, giving ${\sum }_{n\in S}{n}^{-s}=(1-2^{-ϵ})\mathrm{/}[ϵ(1-{10}^{-ϵ})]+O(1)$. Since $ϵ\zeta (1+ϵ)\to 1$, the relative density is $\log 2\mathrm{/}\log 10$.

For primes, the Euler logarithm gives ${\sum }_p{p}^{-s}=\log (1\mathrm{/}ϵ)+O(1)$: all terms of exponent at least two sum to at most ${\sum }_{n\ge 2}{\sum }_{j\ge 2}{n}^{-j}<\mathrm{\infty }$, and the zeta pole determines its logarithm. Finite sets have zero relative density, and finite additivity for disjoint sets follows by taking limits of their finite sum identities.

The qualitative prime number theorem and $\mathrm{Li}(x)\sim x\mathrm{/}\log x$ give $E(x):=\pi (x)-\mathrm{Li}(x)=o(x\mathrm{/}\log x)$; the Li asymptotic follows by integration by parts and splitting its integral at $\sqrt{x}$. Put $a_k={10}^k$, $b_k=2a_k$, and ${\eta }_k={\sup }_{x\ge a_k}\mathrm{\mid }E(x)\mathrm{\mid }\log x\mathrm{/}x$, so ${\eta }_k\to 0$. For $s=1+ϵ$ and $k\ge 1$, Stieltjes integration by parts on $[a_k,b_k)$ gives ${\sum }_{a_k\le p<b_k}{p}^{-s}-{\int }_{a_k}^{b_k}\frac{x^{-s}}{\log x}dx=O\text{\hspace{0.17em}⁣}\left(\frac{{\eta }_k{e}^{-ϵk\log 10}}{k}\right)\mathrm{.}$ Indeed each endpoint term is at most ${\eta }_k{x}^{-ϵ}\mathrm{/}\log x$, and the interior term is bounded by $s{\eta }_k{\int }_{a_k}^{b_k}{x}^{-1-ϵ}\mathrm{/}\log xdx$, of the same order for $1<s\le 2$. Summing over $k$ and absorbing the initial block yields ${\sum }_{p\in S}{p}^{-s}={\int }_2^{\mathrm{\infty }}w(\log x)\frac{x^{-s}}{\log x}dx+o\text{\hspace{0.17em}⁣}(\log \frac{1}{ϵ}),$ because ${\sum }_{k\ge 1}{\eta }_k{e}^{-ϵk\log 10}\mathrm{/}k=o(\log (1\mathrm{/}ϵ))$: split at a fixed large $K$ and use ${\eta }_k\to 0$ on the tail. Here $w$ is the periodic indicator of $[0,\log 2)$ modulo $\log 10$.

Up to an endpoint error at most one, the counts at ${10}^m$ and $2\cdot {10}^m$ are respectively $({10}^m-1)\mathrm{/}9$ and $({10}^{m+1}-1)\mathrm{/}9$. Dividing by the two endpoints yields $1\mathrm{/}9$ and $5\mathrm{/}9$. Thus an integer counting asymptotic already fails despite the density limit.

Set $y=\log x$. The primitive of $w(y)-d$ is bounded because its integral over a period is zero. Integration by parts against $e^{-ϵy}\mathrm{/}y$ therefore bounds its contribution by $O(1)$, uniformly as epsilon tends to zero. The mean contribution is $d{\int }_{\log 2}^{\mathrm{\infty }}{e}^{-ϵy}dy\mathrm{/}y=d\log (1\mathrm{/}ϵ)+O(1)$: split at $1\mathrm{/}ϵ$ and substitute $ϵy$ on the tail. The $o(\log (1\mathrm{/}ϵ))$ error in step 1.3 vanishes after division by the prime denominator from step 1.2, proving the claimed relative Dirichlet density.

For the prime counting limits, fix $J$. At $X={10}^m$, the last $J$ completed blocks, indexed $k=m-j$ with $1\le j\le J$, have counts $\pi (2\cdot {10}^k)-\pi ({10}^k)\sim {10}^k\mathrm{/}\log ({10}^k)$. Relative to $\pi (X)\sim X\mathrm{/}\log X$, their limits sum to ${\sum }_{j=1}^J{10}^{-j}$. Earlier blocks contribute at most $\pi ({10}^{m-J})\mathrm{/}\pi ({10}^m)\to {10}^{-J}$. Let $J$ tend to infinity to get $1\mathrm{/}9$. At $X=2\cdot {10}^m$, the current block contributes relative limit $1\mathrm{/}2$, and the preceding blocks contribute $(1\mathrm{/}2){\sum }_{j\ge 1}{10}^{-j}=1\mathrm{/}18$, totaling $5\mathrm{/}9$. Endpoints are composite for $m$ positive, so they add no ambiguity. These unequal limits refute the inference for primes too.