Lacunary Sequence with positive upper density

A sequence $$\{n_{k}\}$$ is lacunary if $$\forall$$ k $$\in\mathbb{N}$$, $$\frac{n_{k+1}}{n_{k}}$$ $$\geq\lambda\gneq$$ 1.

And natural upper density of a set S $$\subset\mathbb{N}$$ is defined as limsup$$_{n\rightarrow\infty}$$ $$\frac{| S \cap \{ 1, 2, ... , n \}|}{n}$$.

Only examples of lacunary sequences that I could think of were basically geometric sequences; however, they will have upper density zero. I was wondering if there exists lacunary sequences with positive upper density.

• You can have lacunary sequences much sparser than geometric sequences. $n_k = 10^{k!}$, is somewhat well known. Perhaps you mean "[O]nly examples of lacunary sequences that might have positive upper density that I could think of were basically geometric sequences; ..." – Eric Towers May 20 at 21:33
• (1) Title says density, body says upper density. (2) Why the weird symbol $\gneq$, how is that different from $\gt$? (3) In your definition of natural upper density, is $A=S$? – bof May 20 at 22:31

Lacunary sequences cannot have positive upper density. To see this, note that if $$\lambda > 1$$ is such that $$n_{k+1} / n_{k} > \lambda$$ for all $$k$$., then there are at most $$\log(n) / \log(\lambda)$$ elements in the sequence between $$1$$ and $$n$$.
Consider the sequence in log space. Then, note that if there are more than $$\log(n) / \log(\lambda)$$ elements in $${1,...,n}$$, then by the pigeonhole principle, there must be a $$k$$ such that $$\log(n_{k+1}) - \log(n_k) < \log(\lambda)$$ which implies that $$n_{k+1} > \lambda n_{k}$$. A contradiction.