> [!theorem] > > Let $\seq{x_n}$ be a non-negative, non-decreasing [[Sequence|sequence]]. If there exists $a > 0$ and $\varepsilon > 0$ such that > $ > \frac{x_n}{an} > 1 + \varepsilon > $ > infinitely often. Then there exists a *lacunary* sequence $\seq{n_k}$ such that $\frac{x_{n_k}}{a{n_k}} > 1 + \varepsilon/3$ infinitely often. > > On the other hand, if > $ > \frac{x_n}{an} < 1 - \varepsilon > $ > then there exists a lacunary sequence $\seq{n_k}$ such that $\frac{x_{n_k}}{a_{n_k}} < 1 - \varepsilon/3$ infinitely often. > > *Proof*. Let $n_k = \lceil (1 + \varepsilon/2)^k \rceil$, then $n_k$ is a lacunary sequence where > $ > 1 + \frac{1}{3}\varepsilon < \frac{n_{k + 1}}{n_k} < 1 + \frac{2}{3}\varepsilon > $ > eventually. Since $x_n/an > 1 + \varepsilon$ infinitely often, there exists $n$ large enough such that the above holds while $x_n > an(1 + \varepsilon)$. Choose $k \in \nat$ such that $n \in (n_{k-1}, n_{k}]$, then > $ > \begin{align*} > x_{n_k} \ge x_n > an(1 + \varepsilon) &> an_{k - 1}(1 + \varepsilon) \\ > &> \frac{a(1 + \varepsilon)n_k}{1 + \frac{2}{3}\varepsilon} \\ > &= an_k\paren{1 + \frac{\varepsilon}{3 + 2\varepsilon}} > \end{align*} > $ > On the other hand > $ > \begin{align*} > x_{n_k} &\le x_n \\ > &\le an(1 - \varepsilon) \\ > &< an_{k - 1}(1 - \varepsilon)(1 + 2\varepsilon/3) \\ > &< an_{k - 1}\paren{1 - \frac{\varepsilon}{3}} > \end{align*} > $