# Can we have a power density but not a natural density?

For $M \subset \mathbb{N}$ (in this post I follow the convention $\min \mathbb{N} = 1$) and $\alpha \in [0,1]$ define

$$S_{M,\alpha}(x) = \sum_{\substack{n\in M \\ n \leqslant x}} \frac{1}{n^{\alpha}}$$

on $[1,+\infty)$. For sets $A \subset B \subset \mathbb{N}$ with $S_{B,\alpha}(x) \to +\infty$ as $x\to +\infty$, we define the upper $\alpha$-density of $A$ in $B$ as

$$\overline{D}_{\alpha}(A;B) = \limsup_{x\to +\infty} \frac{S_{A,\alpha}(x)}{S_{B,\alpha}(x)}.$$

The lower $\alpha$-density $\underline{D}_{\alpha}(A;B)$ is defined analogously (using $\liminf$ instead of $\limsup$). We say that $A$ has an $\alpha$-density in $B$ if $\overline{D}_{\alpha}(A;B) = \underline{D}_{\alpha}(A;B)$, and then denote that value with $D_{\alpha}(A;B)$. For $\alpha = 0$ we have the familiar natural density (also called asymptotic density) and for $\alpha = 1$ the logarithmic density. The cases $\alpha \in (0,1)$ shall be called "power densities".

Via summation by parts, it is straightforward to show that when $B$ is substantial, i.e. $\lim\limits_{x\to +\infty} S_{B,1}(x) = +\infty$,

$$\underline{D}_{\alpha}(A;B) \leqslant \underline{D}_{\beta}(A;B) \leqslant \overline{D}_{\beta}(A;B) \leqslant \overline{D}_{\alpha}(A;B)$$

for all $A \subset B$ and $0 \leqslant \alpha < \beta \leqslant 1$. So if $A$ has an $\alpha$-density in $B$ for some $\alpha < 1$, then $D_{\beta}(A;B)$ exists for all $\beta \in [\alpha,1]$ and all these densities coincide. And for example the set $A_1$ of positive integers whose first (decimal) digit is $1$ has logarithmic density $\frac{\log 2}{\log 10}$ in $\mathbb{N}$, but $\underline{D}_{\alpha}(A_1;\mathbb{N}) < \overline{D}_{\alpha}(A_1;\mathbb{N})$ for all $\alpha \in [0,1)$.

So the question arises whether there is a substantial set $B$ and a subset $A$ such that $A$ has no natural density in $B$, but $D_{\alpha}(A;B)$ exists for some $\alpha \in (0,1)$.

Again via summation by parts it is easy to see that for well-behaved $B$ - e.g. $\underline{D}_0(B;\mathbb{N}) > 0$, or such that $\frac{S_{B,0}(x)}{x}$ tends to $0$ in a nice fashion, like for the set of primes - the existence of $D_{\alpha}(A;B)$ for an $\alpha < 1$ implies the existence of $D_0(A;B)$.

But I did not find a proof of that for arbitrary substantial $B$, nor did I manage to find an example where $D_{\alpha}(A;B)$ exists for some $\alpha \in (0,1)$ and $D_0(A;B)$ doesn't exist. So:

Do there exist $A \subset B \subset \mathbb{N}$ such that $B$ is substantial, $A$ doesn't have a natural density in $B$ and $D_{\alpha}(A;B)$ exists for an $\alpha \in (0,1)$?

• Why do you exclude $\alpha = 1$ from the boxed question? – mathworker21 Nov 11 '17 at 2:10
• The case $\alpha = 1$, the logarithmic density, is no problem. It's easy to find such examples. The difficulty is with $D_{\alpha}$ for $0 < \alpha < 1$. Of course such an $A$ must have logarithmic density in $B$, but the problem is to get below $\alpha = 1$ without having a natural density. – Daniel Fischer Nov 11 '17 at 11:43

## Yes, we can!

For $k \in \mathbb{N}$, let $a_k = \bigl((2k-1)!\bigr)^2$, $b_k = \bigl((2k)!\bigr)^2$ and $c_k = (1 + \varepsilon_k)a_k$, where $\varepsilon_1 = 1$ and $\varepsilon_k = \frac{1}{k\log k}$ for $k > 1$, and

$$B = \bigcup_{k = 1}^{\infty} [a_k, b_k),\quad A = \bigcup_{k = 1}^{\infty} [a_k, c_k)\,.$$

Then $A$ - and a fortiori $B$ - is substantial:

$$\sum_{k = 1}^K \Biggl(\sum_{a_k \leqslant n < c_k} \frac{1}{n}\Biggr) > \sum_{k = 1}^K \log (1 + \varepsilon_k) > \frac{1}{2}\sum_{k = 1}^K \varepsilon_k > \frac{1}{4} \log \log K\,.$$

$A$ has no natural density in $B$:

For $k \geqslant 2$ we have $S_{A,0}(c_k) \geqslant \varepsilon_k a_k$ and $S_{B,0}(c_k) \leqslant b_{k-1} + \varepsilon_k a_k < \frac{k+\log k}{k^2\log k} a_k$, so

$$\frac{S_{A,0}(c_k)}{S_{B,0}(c_k)} \geqslant \frac{k}{k+\log k}$$

and hence $\overline{D}_0(A;B) = 1$. But $\underline{D}_0(A;B) = 0$ since

$$\frac{S_{A,0}(b_k)}{S_{B,0}(b_k)} \leqslant \frac{(1+\varepsilon_k)a_k}{b_k - a_k} = \frac{1+\varepsilon_k}{4k^2-1}\,.$$

And we have $D_{\frac{1}{2}}(A;B) = 0$ since for $a_{k+1} \leqslant m < a_{k+2}$

$$S_{B,\frac{1}{2}}(m) \geqslant \sum_{n = a_{k}}^{b_{k}-1} \frac{1}{\sqrt{n}} > 2\sum_{n = a_k}^{b_k-1} \frac{1}{\sqrt{n+1} + \sqrt{n}} = 2(\sqrt{b_k} - \sqrt{a_k}) = (4k-2)\cdot (2k-1)! \geqslant (2k)!$$

and

$$S_{A,\frac{1}{2}}(m) \leqslant \sum_{n = 1}^{c_k-1} \frac{1}{\sqrt{n}} + \sum_{n = a_{k+1}}^{c_{k+1}-1} \frac{1}{\sqrt{n}} < 2\sqrt{c_k} + \varepsilon_{k+1} \sqrt{a_{k+1}} < 3(2k-1)! + \frac{(2k+1)!}{(k+1)\log (k+1)}$$

whence

$$\frac{S_{A,\frac{1}{2}}(m)}{S_{B,\frac{1}{2}}(m)} < \frac{3}{2k} + \frac{2}{\log (k+1)}\,.$$

It is not hard to modify the example to get $0 < D_{\frac{1}{2}}(A';B) < 1$.