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Notation: Let $K$ be a subset of natural numbers $\mathbb{N}$. We set $$\delta(K)=\lim\limits_{n\rightarrow \infty}\frac{1}{n}|\{k\in K:k\leq n\}|.$$

Question: Let $(a_{n})_{n\in \mathbb{N}}$ be a sequence of reals such that $\lim\limits_{n\rightarrow \infty}a_{n}=0$. Is there a set $K=\{k_{j}:j\in \mathbb{N}\}\subseteq\mathbb{N}$ with $\delta(K)=1$ such that $$|a_{k_{j}}|\leq \frac{1}{j^{3}},\quad \forall j\in \mathbb{N} ?$$

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  • $\begingroup$ Are there nice characterisations of subsets $K$ for which this $\delta(K)$ exists? $\endgroup$
    – Arnaud Mortier
    Commented Dec 30, 2017 at 14:33
  • $\begingroup$ It seems that there is no nice characterization. $\endgroup$
    – Dongyang Chen
    Commented Dec 31, 2017 at 1:26

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No, this does not hold in general. If we simply let $a_n=\frac1n$, then the given assumptions imply $\delta(K)=0$.

Note that each $k_j\ge j^3$, hence on any interval $[1,n]$ with $j^3\le n< (j+1)^3$ there are at most $j$ many elements of $K$. So for all $n$, $$\frac1n|\{k\le n:k\in K\}|\le \frac{j}n\le\frac{\sqrt[3]n}n\to 0.$$

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  • $\begingroup$ That's great! Thanks, Kjos-Hanssen. $\endgroup$
    – Dongyang Chen
    Commented Dec 31, 2017 at 1:25

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