Problem: Let $X_1,X_2,\dots$ be an i.i.d. sequence of $\text{Bernoulli}$ random variables. Let $L_n$ be the length of the longest consecutive sequence of ones among $X_1,\dots,X_n$. Show that $$P\left(\limsup\limits_{n\to\infty}\frac{L_n}{\log_2(n)}\leq1\right)=1.$$ Attempt: In a hint to the problem it has been suggested that it suffices to show that for any $\varepsilon>0$ we have $P(L_n>(1+\varepsilon)\log_2(n)\text{ i.o})=0$. Further, it has been pointed that to prove the above we can use the Borel-Cantelli lemma with the events $$A_n=\left\{X_n=X_{n-1}=\cdots=X_{n-\lfloor(1+\varepsilon)\log_2(n)\rfloor}=1\right\}.$$ Therefore, we need to compute the probability of the event $A_n$ above. Using independence we have $$P(A_n)=P(X_n=1)\cdot P(X_{n-1}=1)\cdots P(X_{n-\lfloor(1+\varepsilon)\log_2(n)\rfloor}=1)=\frac{1}{2\cdot2^{\lfloor(1+\varepsilon)\log_2(n)\rfloor}}.$$ Then we have $$\sum_{n=1}^\infty P(A_n)\leq\sum_{n=1}^\infty \frac{1}{2\cdot2^{(1+\varepsilon)\log_2(n)}}=\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n^{1+\varepsilon}}<\infty,$$ where in the last step we used the fact that $\sum_{n=1}^\infty\frac{1}{n^p}<\infty$ whenever $p>1.$ It follows from the Borel-Cantelli lemma that $$P(\omega\,:\,\omega\in A_n\text{ for infinitely many }n)=0.$$ This implies that $$P(L_n>(1+\varepsilon)\log_2(n)\text{ i.o.})=0.$$ This in turn implies that $$P\left(\frac{L_n}{\log_2(n)}\leq(1+\varepsilon)\text{ for all but finitely many }n\right)=1.$$ Taking the limit superior for all such $\omega$ in the event above we have that $$P\left(\limsup\limits_{n\to\infty}\frac{L_n}{\log_2(n)}\leq(1+\varepsilon)\right)=1,$$ and since $\varepsilon>0$ was arbitrary, we have that $$P\left(\limsup\limits_{n\to\infty}\frac{L_n}{\log_2(n)}\leq1\right)=1.$$
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