# 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.

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.

• how did you get the last bound? Shouldn't $\max_{k \in ((n-1)^4,n^4]} |S(k) - S((n-1)^4)|=\sum_{k=(n-1)^4}^{n^4}1/k \sim 4\log((n-1)/n)$? – Hasse1987 Aug 3 '17 at 17:20
• also I'm missing where you use that the $X_n$ are centered? – Hasse1987 Aug 3 '17 at 18:48
• @Hasse1987, 1. I bound this sum by numbers of terms (which I overestimate by $4n^3$) times the largest term (which is less that $(n-1)^{-4}$). This gives the same asymptotics: $4n^3/(n-1)^4 \sim 4/n\sim 4\log(1-1/n)$. 2. When I write $\mathrm{E}[(S(m)-S(l))^2] = \sum_{k=l+1}^m \mathrm{E}[X^2_k/k^2]$, I use that $\mathrm{E}[X_i X_j] = 0$ for $i\neq j$. – zhoraster Aug 4 '17 at 6:09
• @Hasse1987, 1. $\mathrm{E}[X_iX_j]=\mathrm{E}[X_i]\mathrm{E}[X_j]=0$; here I use assumptions of zero correlation and zero mean. 2. Yes, you can consider this as a consequence of the monotone convergence theorem. – zhoraster Aug 4 '17 at 7:18