How to prove $\lim\sup(ks_n)=k\cdot \lim\sup s_n$ if $s_n$ is bounded sequence and $k$ is nonnegative real number?
I think we need to prove first that $\sup\{ks_n:n>N\}=k\sup\{s_n:n>N\}$ and then we can take the limit. But how do we do the first step?
If $k=0$ the claim is straightforward.
Let $k>0$. For a fixed $N\in\mathbb{N}$ let $\sigma := \sup\{s_n: n > N\}$. Since $(s_n)$ is bounded we clearly have $\sigma\in\mathbb{R}$.
By definition of $\sup$, for every $\epsilon > 0$ there exists an index $m>N$ such that $\sigma - \epsilon / k < s_m \leq \sigma$, hence $k\sigma - \epsilon < k s_m \leq k\sigma$. This proves that $\sup\{k s_n:\ n>N\} = k\sigma$.