Durrett Theorem 2.3.9 Proof I am trying to understand the proof of Thm 2.3.9 in Durrett's Probability Theory and Examples (5th edition). The theorem and proof are here:


I am not sure why it is enough to show $\frac{ET_{k+1}}{ET_{k}}\rightarrow 1$ to conclude that $\frac{S_{n}}{E(S_{n})}\rightarrow 1$ a.s.?
Any help will be appreciated.
 A: At the point in the proof where Durrett says "it is enough to show $ET_{k+1}/ET_k\to 1$," we have already proved that $T_k/ET_k\to 1$ almost surely. Since $T_{k+1}/ET_{k+1}$ is a subsequence of the a.s. convergent sequence $T_k/ET_{k}$, it too converges a.s. to the same limit. So we are not claiming that $ET_{k+1}/ET_k\to 1$ implies that $T_k/ET_k\to 1$ a.s., but we will use $ET_{k+1}/ET_k\to 1$ together with $T_k/ET_k\to 1$ a.s. to finish the proof.
Let $\omega\in \Omega$ be such that $\lim_{k\to\infty}T_k(\omega)/ET_k=1$ (i.e., the numerical sequence $T_k(\omega)/ET_k\to 1$ as $k\to\infty$). The set of all such $\omega$ has probability $1$ by what Durrett has already shown. So if we can show that $S_n(\omega)/ES_n\to 1$ for all of these $\omega$, then we will have shown therefore that $P(S_n/ES_n\to 1) \ge P(T_k/ET_k\to 1) = 1$, so $S_n/ES_n\to 1$ a.s.
Now, for such an $\omega$, Durrett writes that we have:
$$
\frac{ET_k}{ET_{k+1}}\cdot\frac{T_k(\omega)}{ET_k} \le \frac{S_n(\omega)}{ES_n} \le \frac{T_{k+1}(\omega)}{ET_{k+1}}\cdot\frac{ET_{k+1}}{ET_{k}}
$$
Before explaining the details of why showing $ET_{k+1}/ET_k\to 1$ implies that $S_n(\omega)/ES_n\to 1$, I want to give you the intuition for why this is true. If $ET_{k+1}/ET_k\to 1$, then $ET_k/ET_{k+1}\to 1$ (this is an example of the "rule" $a_n/b_n\to 1$ if and only if $b_n/a_n\to 1$). So, since we assumed $T_k(\omega)/ET_k\to 1$ (and therefore the subsequence $T_{k+1}(\omega)/ET_{k+1}\to 1$), if we let $k\to\infty$, the left- and right-hand sides converge to $1$, so if $n\to\infty$ as $k\to\infty$, then the middle term $S_n(\omega)/ES_n\to 1$, as desired.
We have to make precise that $n\to \infty$ as $k\to\infty$ to turn the last paragraph into a proof.
Recall at this point that $T_k = S_{n_k}$, and that $n_k \le n < n_{k+1}$. By its definition, $n_k = \inf\{n : ES_n\ge k^2\}$ is an increasing sequence that tends to $\infty$ as $k\nearrow \infty$ because $ES_n\to\infty$ as $n\to\infty$. Therefore, as $k\nearrow \infty$, we have $n\ge n_k \nearrow \infty$. So we see that, by the definition of $\liminf$ and $\limsup$, we have
$$
1 = \lim_{k\to\infty}\frac{ET_k}{ET_{k+1}}\cdot\frac{T_k(\omega)}{ET_k} \le \liminf_{n\to\infty}\frac{S_n(\omega)}{ES_n} \le \limsup_{n\to\infty}\frac{S_n(\omega)}{ES_n} \le \lim_{k\to\infty}\frac{T_{k+1}(\omega)}{ET_{k+1}}\cdot\frac{ET_{k+1}}{ET_{k}} = 1,
$$
so $\lim_{n\to\infty}S_n(\omega)/ES_n = 1$. We could also see that we have the claim by applying either the definition of the limit of a sequence indexed by $n$, or by using the squeeze theorem.
