Convergence of random harmonic series without $3$ series theorem In Rota's unfinished probability book, problem $57$ on page $4.67$ (pdf page $236$) is to show that the series $\sum_1^\infty X_n/n$ converges with probability $1$, where $X_n$ equals $1$ or $-1$ with equal probability. This occurs early on in the text before anything like the Kolmogorov $3$ series theorem has been proven, so I am looking for more elementary proofs. Kolmogorov's $0-1$ law has been proven, and he notes in the text before the exercise that the event of convergence is a tail event, so that is a likely approach (though I still didn't get far). 
The book is here.
 A: Denote $S(n) = \sum_{k=1}^n \frac{X_n}{n}$. 
Write 
$$
\mathrm{E}\left[\sum_{n=2}^\infty n^3\big(S(n^4) - S((n-1)^4)\big)^2\right] = \sum_{n=2}^\infty n^3\mathrm{E}\left[\big(S(n^4) - S((n-1)^4)\big)^2\right] \\
= \sum_{n=2}^\infty  n^3\sum_{k=(n-1)^4 +1}^{n^4} \mathrm{E}\left[\frac{X_k^2}{k^2}\right]\le  \sum_{n=2}^\infty n^3\cdot \frac{4n^3}{(n-1)^8}<\infty.
$$
Therefore, 
$$
\sum_{n=2}^\infty n^3\big(S(n^4) - S((n-1)^4)\big)^2<\infty
$$
almost surely, in particular, 
$$
n^{3/2}\big|S(n^4) - S((n-1)^4)\big|\to 0, \ n\to\infty,
$$
almost surely. 
Therefore, 
$$
\sum_{n=2}^\infty \big|S(n^4) - S((n-1)^4)\big|<\infty
$$
almost surely. 
On the other hand,
$$
\max_{k \in ((n-1)^4,n^4]} |S(k) - S((n-1)^4)|\le \frac{4n^3}{(n-1)^4}\to 0,n\to\infty.
$$
It follows from the last two facts (an exercise for you) that the sequence of partial sums is Cauchy almost surely. Consequently, it converges almost surely. 
This argument works for any uncorrelated centered variables $X_n$ bounded by the same constant.

As a side remark, I consider the linearity of expectation the most powerful tool in the probability theory. It is a miracle how many things you can prove with it. For example, the above argument uses no other probabilistic tools, just linearity of expectation; everything else is some routine calculus.
