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](https://link.springer.com/article/10.1007/BF02807252), 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](https://en.wikipedia.org/wiki/Davenport%E2%80%93Erd%C5%91s_theorem) theorem states the equivalence of this two notions of densities for sets. See [this post](https://math.stackexchange.com/questions/639136/natural-density-and-logarithmic-density) 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



Logarithmic vs sequential density of a sequence
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 $0$.


