Strong Law Without Summable Tail Probabilities Suppose that $X_{k}$ are independent but not necessarily identically distributed.  Let $S_{n} = \sum_{k=1}^{n}X_{k}$.  Does there exist an example where $\frac{1}{n}S_{n}\rightarrow 0$ almost surely, but $\sum_{n=1}^{\infty}\mathbb{P}(|\frac{1}{n}S_{n}| > \varepsilon)=\infty$ for sufficiently small $\varepsilon$?
 A: Edit: Sorry, I forgot the requirement for the indipendence of the $X_k$. I will leave this answer here anyways, I might help others. But note that it's not the answer that the author was looking for.
The standard example for a sequence that converges almost surely, but not completely, is given by $$X_n = \mathbb{1}_{[0,\frac{1}{n}]}.$$ on the probability space $[0,1]$ with Lesbesgue measure and Borel sigma-algebra.
My idea is to take $$S_n = \sum_{k=1}^n  X_k.$$
Claim 1: $\frac{1}{n} S_n \rightarrow 0$ a.s.
Proof: Let $x \in [0,1]$. Pick $N$ s.t. $\frac{1}{N} < x$. Then
$$\frac{1}{n} S_n(x) = \frac{1}{n} \sum_{k=1}^{N-1} X_k(x) + \frac{1}{n} \sum_{k = N}^n X_k(x).$$
But the seconds summand is always zero due to the choice of $N$ and the first $\to0$ if $n \to \infty$.
Claim 2: $\sum_{n=1}^\infty \mathbb{P}(\frac{1}{n} S_n > \epsilon) = \infty$ for small $\epsilon$.
Proof: Take any $\epsilon < 1$.
Note that $\frac{1}{n}S_n = 1$ on the interval $[0, \frac{1}{n}]$, i.e. on this interval, it will always be $> \epsilon$.
$$ \sum_{n=1}^\infty \mathbb{P}(\frac{1}{n} S_n > \epsilon) \geq  \sum_{n=1}^\infty \mathbb{P}([0, \frac{1}{n}]) = \sum_{n=1}^\infty \frac{1}{n}= \infty. $$
This finishes the proof.
I just came up with this example myself, do you think it's correct?
A: This is the answer suggested in a hint in the comments.  Let $X_{k}$ be independent and equal to $A_{k}$ with probability $p_{k}$ and $0$ otherwise.  Assume that $\sum_{k=1}^{\infty}p_{k} < \infty$ so that $S_{n}$ is almost surely bounded by Borel Cantelli.  Suppose that $A_{k}$ grows quickly enough that, for some fixed $c\in (0,1)$, when $k \ge \lfloor cn \rfloor$ it holds that $A_{k} \ge n$.  Then we obtain the following lower bound on the tail of $\frac{1}{n}S_{n}$ for any $\varepsilon < 1$,
$$
\mathbb{P}(\frac{1}{n}|S_{n}| > \varepsilon)\ge \mathbb{P}(\text{some $X_{k} > 0$ for $k > \lfloor cn \rfloor$}) \ge \sum_{k\ge \lfloor cn \rfloor}^{}p_{k} - \sum_{k_{1},k_{2}\ge \lfloor cn \rfloor}^{}p_{k_{1}}p_{k_{2}}\,.
$$
Choose $p_{k} = k^{-2}$, $c=\frac{1}{2}$, and $A_{k}=2k$.  Applying the lower bounds above and using comparison with integrals of $1/x^{2}$, we see that $\mathbb{P}(\frac{1}{n}|S_{n}| > \varepsilon)\ge (\frac{1}{c}-1)\frac{1}{n} = \frac{1}{n}$.   Therefore, $\sum_{k}^{}\mathbb{P}(\frac{1}{k}|S_{k}| > \varepsilon)=\infty$.
