Is it true that $\limsup a_n \geq \limsup b_n$ implies $\limsup a_n^{s_n} \geq \limsup b_n^{s_n}$ for $a_n, b_n \geq 0$ and $s_n > 0$ and increasing?

Question

Potential first step: If $\limsup a_n \geq \limsup b_n$, along some subsequence and for large enough $n_k$ we have $a_{n_k} \geq b_{n_k}$. Then, $a_{n_k}^{s_{n_k}} \geq b_{n_k}^{s_{n_k}}$, so somehow $\limsup a_{n_k}^{s_{n_k}} \geq \limsup b_{n_k}^{s_{n_k}}$?

Answer 1

Let $a_n = 1-{1 \over n}, b_n = 1+ { 1 \over n}$. Let $s_n = n$.

Then $1 = \lim\sup_n a_n \ge \limsup_n b_n = 1$, but ${1 \over e} = \lim\sup_n a_n^{s_n} < \limsup_n b_n^{s_n} = e$.