Could one generalize this result in the following way?
Let $(a_n)_{n \in \mathbb{N}}$ be a nondecreasing sequence of positive real numbers. Then for any strictly increasing subsequence $n_k$ with $\lim_{k \to \infty} n_k = \infty$
$\limsup_{n \to \infty} \frac{a_n}{n} = \limsup_{k \to \infty} \frac{a_{n_k}}{n_k}$
I think the proof proposed there already would hold in this more general setting, but I'm not sure if I'm missing some special cases in which the result above doesn't hold. Could someone share his/her thoughts?
Let $a_n=2^k$ for $2^k\le n< 2^{k+1}.$ Then $a_n\le n$ and $a_{2^k}=2^k.$ Hence $$\limsup {a_n\over n}=1.$$ We have $a_{2^{k+1}-1}=2^k.$ Therefore $$\lim_k {a_{2^{k+1}-1}\over2^{k+1}-1}={1\over 2}$$ Hence the conclusion fails.
Remark I have impression that the conclusion may hold under additional assumption $$\lim_k{n_{k+1}-n_k\over n_k-n_{k-1}}$$