I'm quiet new to the asymptotic world, so apology in advance if this question seems too trivial for you experts.
Given $\frac{2kn2^{-k}}{E[X]}.$ As $k \sim 2\text{log}_{2} n$, the numerator is $n^{-1+o(1)}.$
I don't understand how the numerator comes out to be $n^{-1+o(1)}.$ Using $k \sim 2\text{log}_{2} n$,
$2kn2^{-k} = \frac{2kn}{2^{k}} \sim \frac{4n\text{log}_{2}n}{n^{2}} = \frac{4\text{log}_{2} n}{n}$.
For the last term to be $n^{{-1}+o(1)}$, $\forall \epsilon >0$, there exists $M$ such that for $n>M$, $\frac{4\text{log}_{2} n}{n} < n^{{-1}+\epsilon}=\frac{n^{\epsilon}}{n} \implies 4\text{log}_{2} n<n^{\epsilon}$.
As $n \rightarrow \infty, 4\text{log}_{2} n \rightarrow \infty $ but $n^{\epsilon} \rightarrow 1$, so how could there exist such $M$?
Thanks you!