Let $\sum a_n$ be a conditionally convergent sum of real numbers, and $\epsilon_n$ a sequence of independent identically distributed Bernoulli random variables with $\epsilon_n = 1$ or $-1$ with probability $\frac{1}{2}$ each.

Is it true that $\sum \epsilon_n a_n$ converges almost surely?

  • 3
    $\begingroup$ I think the answer is "no" for $a_n = \frac{(-1)^n}{\sqrt{n}}$. See Section 3 of these notes of (frequent MO contributor) Pete L. Clark: alpha.math.uga.edu/~pete/UGAVIGRE08.pdf; in particular, look at Theorem 10 there. $\endgroup$ Commented Jan 16 at 0:55
  • $\begingroup$ @SamHopkins I just so happened to be thinking about lots of these kind of rearrangement problems for series this morning, so this paper is a gold mine! $\endgroup$
    – Nate River
    Commented Jan 16 at 0:57
  • $\begingroup$ I posted this as a comment and not an answer because I am not an analyst and I just found that pdf via some educated googling. But if it answers your question I am happy to post it as an answer. $\endgroup$ Commented Jan 16 at 0:58
  • $\begingroup$ @SamHopkins That would be great. I was just about to ask about Theorem 10 as well, about the necessary and sufficient conditions for almost sure convergence (since by Kolmogorov there are only two possibilities, convergence or divergence almost surely). Maybe you could add that in the answer as a further remark? $\endgroup$
    – Nate River
    Commented Jan 16 at 0:59

2 Answers 2


The answer is no, $\sum \epsilon_n a_n$ need not almost surely converge. For instance, with $a_n = \frac{(-1)^n}{\sqrt{n}}$, the random series $\sum \epsilon_n a_n$ converges with probability zero.

This follows from a very precise theorem of Rademacher-Paley-Zygmund, which says that the following are equivalent for $\{a_n\}$ any sequence of real numbers:

  1. The probability that $\sum \epsilon_n a_n$ converges is $1$.
  2. The probability that $\sum \epsilon_n a_n$ is bounded is $1$.
  3. $\sum a_n^2 < \infty$.

(Of course, to conclude that the series converges with probability $0$ in the alternative case, we appeal to Kolmogorov's $0,1$-Law.) I found this theorem stated as Theorem 10 in the notes "Almost sure limit sets of random series" by Pete L. Clark.


For the question at hand, we don't need the full strength of the RPZ theorem cited by Sam Hopkins; just the Kolmogorov three-series theorem. Since $\sum a_n$ converges (conditionally), we have $a_n \to 0$ and in particular the sequence $|a_n|$ is (strictly) bounded by some $A$. Using this $A$ in the three-series theorem with $X_n = \epsilon_n a_n$, we have $Y_n = X_n$. Then:

  • $P(|X_n| \ge A) = 0$ for all $n$ so the convergence of $\sum P(|X_n| \ge A)$ is trivial;

  • $E[Y_n] = E[X_n] = 0$ so the convergence of $\sum E[Y_n]$ is also trivial.

Thus $\sum X_n$ converges almost surely if and only if $\sum \operatorname{Var}(Y_n) = \sum a_n^2$ converges.

Of course, it is easy to find series such that $\sum a_n$ converges conditionally but $\sum a_n^2$ diverges, such as $a_n = (-1)^n/\sqrt{n}$ as mentioned by Sam (and also by the Wikipedia page linked above).


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