Some questions about the longest run 

*

*The upper bound follows from Borel-Cantelli lemma,I only know that $$\displaystyle \limsup_{ n\to \infty}\ell_{n}/\log_{2}n\leq 1\quad \text{a.s.}$$
Intuitively,$$\displaystyle \limsup_{ n\to \infty}\ell_{n}/\log_{2}n\leq 1\quad \text{a.s.} \quad\Leftrightarrow\quad \displaystyle \limsup_{ n\to \infty}L_{n}/\log_{2}n\leq 1\quad \text{a.s.}$$
but how can I get this equivalence rigorously?

*For the lower bound, I don't understand that $\text{“we break the first}$ $n$ trials into disjoint blocks of length $$\left [ \left (1-\epsilon \right )\log_{2}n \right ]+1,$$on which the with the variables are all 1 with probability $$2^{-\left [ \left (1-\epsilon \right )\log_{2}n \right ]-1}\quad ."$$And why$$P(L_{n}\leq(1-\epsilon)\log_{2}n)\le (1-n^{-(1-\epsilon)}/2)^{n/(\log_{2}n)}\quad ?$$
 A: Upper bound
Since $L_n\ge \ell_n$, it should be obvious that
$$
\limsup_{ n\to \infty}\ell_{n}/\log_{2}n\leq 1\quad \text{a.s.} \quad\impliedby\quad \displaystyle \limsup_{ n\to \infty}L_{n}/\log_{2}n\leq 1\quad \text{a.s.}
$$
To get the forward implication, assume $\limsup_{ n\to \infty}\ell_{n}/\log_{2}n\leq 1\; \text{a.s.}$ This mean for each $\epsilon>0$, there exists $N$ such that $n\ge N$ implies $\ell_n\le (1+\epsilon)\log_2 n$. Then for $n\ge N$, we can write
$$
L_n=\max(L_N, \ell_{N+1},\ell_{N+2},\dots,\ell_{n})\le \max(L_N,(1+\epsilon)\log_2 n),
$$
which proves
$$
\frac{L_n}{\log_2 n}
\le \max\left(\frac{L_N}{\log_2 n},1+\epsilon\right)
$$
Since $L_N/\log_2 n\to 0$ as $n\to\infty$, this proves that $L_n/\log_2 n$ is also eventually less than $1+\epsilon$.
Lower Bound
Consider any integer $b$, where $1\le b\le n$. Let us show that
$$
P(L_n \le b)\le (1-2^{-b+1})^{\lfloor n/(b+1)\rfloor}
$$
Note that the event that $\{L_n\le b\}$ implies that
$$
\hspace{2.3cm}X_1,X_2,\dots,X_{b+1}\text{ are not all $1$, and}\\
X_{(b+1)+1},X_{(b+1)+2},\dots,X_{2(b+1)}\text{ are not all $1$, and}\\
X_{2(b+1)+1},X_{2(b+1)+2},\dots,X_{3(b+1)}\text{ are not all $1$, and}\\
\vdots
$$
and so on. This is what is meant by "breaking the trials into disjoint blocks of length $b+1$".
The probability that the first row of variables are all $1$ is $2^{-(b+1)}$, so the probability the first row is not all one is $(1-2^{-b+1})$. The same is true for all rows. The number of rows is $\lfloor n/(b+1)\rfloor$. Multiplying all these probabilities together, the probability of the intersection is therefore $(1-2^{-b+1})^{\lfloor n/(b+1)\rfloor}$.
The author is appyling this result in the case where $b=(1-\epsilon)\log_2 n$.
