Resnick -Probability Path - Exercise 7.40 In this problem, Resnick states that if $\{X_n,n \geq 1\}$ is an independent sequence of random variables and $S_n := \sum_{i=1}^{n} X_n$, then $\frac{S_n}{n} \overset{a.s}{\to} 0$ if and only if $\frac{S_n}{n} \overset{P}{\to} 0$ and $\frac{S_{2^n}}{2^n} \overset{a.s}{\to} 0$
My question is the following: Why did Resnick include the assumption of independence?

I do not believe this is a necessary assumption. The non-immediate direction is to show that $\frac{S_n}{n} \overset{P}{\to} 0$ and $\frac{S_{2^n}}{2^n} \overset{a.s}{\to} 0$ implies $\frac{S_n}{n} \overset{a.s}{\to} 0$. But to do this, we do not need independence. Since $\frac{S_n}{n}$ is cauchy i.p -- by virtue of $\frac{S_n}{n}$ converging in probability -- it is also cauchy a.s (since $\psi_N = \sup_{m,n \geq N} |\frac{S_n}{n} - \frac{S_m}{m}|$ is a monotonically decreasing sequence). And $\frac{S_n}{n}$ being cauchy a.s + $\frac{S_{2^n}}{2^n} \overset{a.s}{\to} 0$ is all we need to conduct a straightforward argument to conclude $\frac{S_n}{n} \overset{a.s}{\to} 0$.
The other direction is an immediate consequence of the definition of almost sure convergence (and does not rely on the independence assumption)
