For every positive integer $n$, let $s_n$ be a uniformly distributed random 01 string with length $n$. For a positive integer $k\leq n$, we say $s_n$ has $k$ consecutive 1's if it contains a length-$k$ substring "$11\dots 11$". Define $L_n$ as the largest $k$ such that $s_n$ has $k$ consecutive 1's. Prove that $\lim_{n\to\infty}\frac{L_n}{\log_2 n}=1$ with probability 1.
I'm only able to show that $\liminf_{n\to\infty}\frac{L_n}{\log_2 n}\geq 1$ with probability 1, by estimating $\mathbb{P}(L_n\leq (1-\varepsilon)\log_2 n)$ for $\varepsilon\in (0,1)$. I have successfully proved that the series $\sum_{n}\mathbb{P}(L_n\leq(1-\varepsilon)\log_2 n)$ converges, then by Borel-Cantelli lemma, $\liminf_{n\to\infty}\frac{L_n}{\log_2 n}\geq 1-\varepsilon$ with probability 1. Since $\varepsilon$ is arbitrary, $\liminf_n \frac{L_n}{\log_2 n}\geq 1$ with probability 1. However, similar argument does not work for the other side. This is because $\sum_{n}\mathbb{P}(L_n\geq(1+\varepsilon)\log_2 n)=\infty$ if $\varepsilon$ is sufficiently small, in which case Borel-Cantelli lemma cannot be applied. How can I bound $\limsup_{n\to\infty} \frac{L_n}{\log_2 n}$?