# Logarithmic vs sequential density of a sequence

Given a sequence of complex numbers $$\{a_n\}_n$$, one says that this sequence admits $$a$$ as a sequential density if $$\underset{N_s\to\infty}{\lim}\frac{1}{N_s}\sum_{n=1}^{N_s} a_n = a$$ where $$N_s = 2^{2^s}$$ for instance. The sequence admits $$a$$ as a logarithmic density if $$\underset{N\to\infty}{\lim}\frac{1}{\log N}\sum_{n=1}^N \frac{1}{n} a_n = a.$$ If the the sequence has a sequential density, does it also have a logarithmic one, with the same limit ?

Motivation: In a paper called Problems of Almost Everywhere convergence related to harmonic analysis and number theory, Jean Bourgain states that for any function $$f$$ in $$L^2(\mathbf{T})$$ the sequence of its Riemann sums $$\{R_nf(x)\}$$ admits $$\int_0^1f$$ as a logarithmic density. In the proof, it turns out that Bourgain reduces this convergence to an $$L^2$$-maximal inequality

$$\left\Vert\underset{N_s=2^{2^s}}{\sup}\frac{1}{N}\left\vert\sum_{n=1}^NR_nf\right\vert\right\Vert_2\leq C \left\Vert f \right\Vert_2$$

which actually proves the sequential density of the sequence $$\{R_nf(x)\}$$ for almost every $$x$$ in $$\mathbf{T}$$.

In number theory, the Davenport-Erdös theorem states the equivalence of this two notions of densities for sets. See this post for instance. I don't know where to find a proof of this theorem in order to adapt it for sequences.

• So I think the answer to your question is "no" unless you have some kind of growth conditions on the $a_n$. For instance if $a_{2^{2^n}}=2^{2^n}$ and $a_{2^{2^n}+1}=-2^{2^n}$ and all other terms are 0, the first sequence has limit 1, whereas the second sequence has limit 0. – Anthony Quas Aug 10 at 16:30
The answer is "no" even if we assume that the $$a_n$$ are bounded. For example, take $$a_n=1$$ if the fractional part of $$\log_2\log_2 n$$ is between $$0$$ and $$\frac12$$ (equivalently, if $$n$$ is between $$2^{2^k}$$ and $$2^{2^{k+1/2}}$$ for some integer $$k$$), and $$a_n=0$$ otherwise. Then the sequential limit will equal $$0$$, but the lim sup of the logarithmic-density expression equals $$1$$ while its lim inf equals $$0$$.