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 $$|a\_{k\_{j}}|\leq \frac{1}{j^{3}},\quad \forall j\in \mathbb{N} ?$$



A question on statistically convergent sequences
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 $n$,
$$\frac1n|\{k\le n:k\in K\}|\le \frac{j}n\le\frac{\sqrt[3]n}n\to 0.$$


