# Equivalence of Dirichlet density and asymptotic density

Let $A, B$ be subsets of the natural numbers with $A \subseteq B$. The asymptotic density of $A$ in $B$ is $$\lim_{N\to \infty}\frac{\text{number of elements of }A\text{ below }N}{\text{number of elements of }B\text{ below }N}.$$ The Dirichlet density is

$$\lim_{s\to 1^+}\frac{\sum\limits_{n\in A}n^{-s}}{\sum\limits_{n\in B}n^{-s}},$$

with $s$ going to $1$ from the right, so the sums are finite. I want to show that if the natural density exists, then the Dirichlet density also exists and is the same. A paper by Bell and Burris proves this in a more general context, but it's a bit involved. Lang says this is "an easy exercise" so I must be missing something. Can someone help me out?

Thanks.

(Edit: I forgot to say that we also require $\sum_{n\in B} n^{-1} = \infty$, otherwise there are obvious counterexamples.)

• Kedlaya gives it as an exercise in his notes, www-math.mit.edu/~kedlaya/18.785/dirichlet.pdf and gives as a hint: use partial summation. – Gerry Myerson Nov 5 '11 at 21:14
• Yep, I saw that too. I assume he means Abel's "summation by parts" lemma. I've tried but I'm always just a little bit short of the result. – Anna B Nov 5 '11 at 22:07

## 1 Answer

The hint is correct.

1. For every $A\subseteq\mathbb N$ and $n\geqslant0$, let $N^A_n$ denote the number of elements of $A$ between $1$ and $n$. Then $$D_A(s)=\sum\limits_{n\in A}n^{-s}=\sum\limits_{n\geqslant1}n^{-s}\cdot(N^A_n-N^A_{n-1})=\sum\limits_{n\geqslant1}(n^{-s}-(n+1)^{-s})\cdot N^A_n.$$
2. Assume $A\subseteq\mathbb N$ and $B\subseteq\mathbb N$ are such that $N_n^A\leqslant uN_n^B+v$ for every $n$, then $$D_A(s)\leqslant\sum\limits_{n\geqslant1}(n^{-s}-(n+1)^{-s})\cdot(uN^B_n+v)=uD_B(s)+v.$$
3. Assume from now on that the asymptotic density of $A\subseteq\mathbb N$ in $B\subseteq\mathbb N$ is $\delta$. Then, for every $u<\delta<u'$, there exists some finite $v$ and $v'$ such that $uN^B_n+v\leqslant N^A_n\leqslant u'N_n^B+v'$ for every $n$, hence $$u+\frac{v}{D_B(s)}\leqslant\frac{D_A(s)}{D_B(s)}\leqslant u'+\frac{v'}{D_B(s)}.$$
4. When $s\to1^+$, $D_B(s)\to D_B(1)$ which is infinite by hypothesis, hence the limit points of the ratio $D_A(s)/D_B(s)$ when $s\to1^+$ are between $u$ and $u'$. Since $u$ and $u'$ may be as close to $\delta$ as one wants, this proves that the Dirichlet density of $A$ in $B$ exists and that it is $\delta$.

Edit The idea of the proof, if there should be one, is to translate the asymptotic hypothesis that two positive sequences $(x_n)_{n}$ and $(y_n)_{n}$ are such that $\lim \frac{x_n}{y_n}=\ell$ into some nonasymptotic inequalities, as follows.

First, by definition of the limit, for every $u<\ell<u'$, $uy_n\leqslant x_n\leqslant u'y_n$ for every $n\geqslant N$. Second, consider $v=-u\max\{y_n\mid n\leqslant N\}$ and $v'=\max\{x_n\mid n\leqslant N\}$ hence $v\leqslant0\leqslant v'$. Then, for every $n\leqslant N$, $uy_n+v\leqslant0$, $v'\leqslant u'y_n+v'$ and $0\leqslant x_n\leqslant v'$, hence $uy_n+v\leqslant x_n\leqslant u'y_n+v'$. Finally, $uy_n+v\leqslant x_n\leqslant u'y_n+v'$ for every $n$.

• Perfect. Thank you very much. – Anna B Nov 5 '11 at 22:25