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.

Thank you in advance