I'm trying to solve Exercise 5.4-8 in the fourth edition of CLRS:
$\star$ 5.4-8
Sharpen the lower bound on streak length by showing that in $n$ flips of a fair coin, the probability is at least $1 - 1/n$ that a streak of length $\lg n- 2\lg \lg n$ consecutive heads occurs.
While there is a closed form solution for the probability of a streak, I found it difficult to use for getting the bound needed here.
I also found a supposed solution (listed as 5.4-7 there), using a similar approach to the text, where the sequence is partitioned into $\lfloor{n/s} \rfloor$ subsequences of length $s$ each. However, I think there are some issues with this solution. For starters, I believe a streak of length $s=\lg n - 2 \lg \lg n$ actually means $s = \lceil\lg n - 2 \lg \lg n \rceil$. Also the floor at the exponent $n/s$ appears to be missing. I tried bounding the floors and ceilings more carefully and ended up with the inequality $$\left(1-\frac{\log _2^2(n)}{2 n}\right){}^{\frac{n}{\log _2(n)-2 \log _2\left(\log _2(n)\right)}-1} \leq \frac{1}{n} $$ which unfortunately fails around $n=2 \times 10^{11}$.
I cannot see how to remedy the proof in the link, nor how to get this bound in any other way. I still believe the statement in the exercise is true, and would greatly appreciate help proving it. Thanks.
