We consider the experiment on tossing a fair coin infinitely many times, and let $T_{n}$ be the even that the $n^{\text{th}}$ coin is a tail. The problem is to determine (here $\left[ x \right]$ denotes the integer nearest to $x$) $$\mathsf{P}\Bigl( \bigcap_{k=1}^{\left[ \log_{2} (n) \right]} T_{k} \text{ infinitely often}\Bigr).$$
Here is what I have tried so far: For every $n \geq 1$, set $B_{n} = \bigcap_{k=1}^{\left[\log_{2}(n)\right]} T_{k}$. Then, since the $T_{k}$'s are independent, we see that $\mathsf{P}(B_{k}) = 1/2^{\left[\log_{2}(k)\right]}$, and thus $\sum_{n\geq 1} \mathsf{P}(B_{n}) = +\infty$. Therefore, a reasonable guess is that the desired probability is $1$. Unfortunately, the $B_{n}$'s are not independent (they have overlap), so we cannot apply Borel-Cantelli lemma directly. This leads me to the following: find a strictly increasing sequence $(k_n)_{n\geq 1}$ such that the $(B_{k_n})_{n}$ do not overlap and satisfies $$ \sum_{n\geq 1} \frac{1}{2^\left[\log_{2}(k_{n})\right]} = +\infty. $$ In particular, in other that the $(B_{k_n})_{n}$ do not overlap, we need \begin{equation} k_{n} + \left[\log_{2}(k_{n})\right] < k_{n+1} +1. \quad (1)\end{equation}
I am aware that one might choose $k_{n} = \left[n \log_{2}(n^2) \right]$; however, we then have to check (1), which does not seem do-able in this case.
Here is my question: Is there a simplier choice for $(k_n)_{n \geq 1}$, i.e., a choice such that we can easily verify (1)?
Any help/hint is highly appreciated.