We flip a fair coin repeatedly and independently. Let $N_n$ be the number of consecutive heads beginning from the $n^{th}$ flip. (For example, $N_n =0$ if the $n^{th}$ flip is a tail, and $N_n =2$ if the $n^{th}$ and $(n+1)^{th}$ flips are heads but the $(n + 2)^{th}$ flip is a tail). Show that $$\limsup_{n\rightarrow\infty}\frac{N_n}{\log n}=\frac{1}{\log 2} \ \ \text{a.s.}.$$
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I suppose that problem like this is to use Borel-Cantelli two times to give an upper and lower bound of LHS, it's easy to obtain that $\mathbb{P}[\limsup_{n\rightarrow\infty}\frac{N_n}{\log n}\ge\frac{1+\epsilon}{\log 2}]=0$ using Borel-Cantelli, but I cannot obtain the other similar inequality since $N_n$s are not independent. What should I do?
Thanks!
Edited: Question was changed to prove the almost sure result.
Edited 2: I have an idea, that is if we can prove that $\mathbb{P}[\sup_{1\le k\le n}\frac{N_k}{\log k}\ge\frac{1}{\log 2}]$ is strictly larger than $0$, then we can (?) apply the Kolmogorov's 0-1 law to obtain that it must be $1$ when $n\rightarrow \infty$? I'm not quite sure if this works······
For the lower bound, the key is to use Borel-Cantelli for some independent terms with let $p_k = \mathbb{P}[N_k \ge \log_2 k]\ge\frac{1}{2k}$. Let $k_1=2, k_{n+1}=k_n+\lceil\log k_n\rceil$, then all $\{p_{k_n}\}$ are independent. Since we know that $\sum_{k=1}^{\infty}\frac{1}{k\log k}=\infty$, we have $\prod_{n=1}^{\infty}(1-p_{k_n})=0$, which is our desired result.
Edit: The proof can actually be seen in Durrett’s book.