The divergence of the harmonic series is familiar: the partial sums of positive integer reciprocals grow without bound. A less familiar but still well-known result is Kempner's series: If we only use integers whose base-10 expansion contains a 9, then the sum of reciprocals converges. The intution this suggests is that the convergence of such sums depend on what fraction of the integers are excluded.

As a demonstration of this, one can consider the following probabilistic version of the harmonic series. Let $\{X_n\}_{n=1}^\infty$ be a sequence of fair coin flips with values $\pm 1$. Then we can ask whether or not the series $\sum_{n=1}^\infty X_n/n$ converges. The answer is provided by Kolmogorov's three-series theorem, which supplies conditions for the convergence of a random series $\sum_{n=1}^\infty X_n$. As discussed in the linked Wikipedia article, the series converges almost surely. (By contrast, replacing $1/n\to 1/\sqrt{n}$ results in almost-sure divergence.)

With that as background, I came up with a different variation on a 'probabilistic' harmonic series. We start with an infinite random sequence $\{X_n\}_{n=1}^\infty$ as before, but now each variable is an iid Poisson variable ($X_n\sim \text{Pois}(\lambda)$). Furthermore, let $Y_n=\sum_{k=1}^n (1+X_k)$. (Note that, beyond a minimum distance of $1$, the gaps between successive integers in this sequence are Poisson-distributed.)

Now consider the series $\sum_{n=1}^\infty Y_n^{-1}$. In more elementary terms, we sum the reciprocals of an integer sequence for which the gaps are Poisson-distributed. What can be said about the probability of convergence for this series? Presumably it again hinges upon use of Kolmogorov's three-series theorem but I'm not familiar enough with such to tackle it myself. (I'd also be happy if references for this case exist.)

  • $\begingroup$ What is $Y_1^{-1}$ when $X_1=0$ (with probability $e^{-\lambda}$)? $\endgroup$
    – Henry
    Apr 21, 2021 at 22:00
  • $\begingroup$ Even if you had $Y_n=\sum\limits_{k=1}^n (X_k+1)$ so making the $Y_n^{-1}$ smaller, I suspect $\mathbb E\left[\sum\limits_{n=1}^m Y_n^{-1}\right] \ge \frac{1}{\lambda+1}\sum\limits_{n=1}^m \frac1n$ and also that divergence to $+\infty$ is almost certain $\endgroup$
    – Henry
    Apr 21, 2021 at 22:08
  • $\begingroup$ Good point. The modification you propose seems reasonable and I'll adopt it. $\endgroup$ Apr 21, 2021 at 22:17
  • $\begingroup$ @Henry If the modified version you propose has an easy resolution, then I'm happy and will accept it. (Of course, I'd want to see "I suspect" made more rigorous.) $\endgroup$ Apr 21, 2021 at 22:22
  • $\begingroup$ With the adaptation, $\mathbb E[Y_n] =n(\lambda+1)$ so $\mathbb E[Y_n^{-1}] \ge \frac1{n(\lambda+1)}$. Meanwhile the probability that a positive integer $x$ is one of the $Y_n$ is bounded below by $e^{-\lambda}$ and tends towards $\frac{1}{\lambda+1}$ so $\sum\limits_{n=1}^\infty Y_n^{-1}$ is not far from being a fraction of an infinite harmonic series $\endgroup$
    – Henry
    Apr 21, 2021 at 22:38

1 Answer 1


Let $A=\{\omega \in \Omega: \frac 1n \sum_{k=1}^n X_k(\omega)\to \lambda  \}$. By the SLLN, $P(A)=1$.

For $\omega\in A$, $Y_n(\omega) = n + n \frac 1n \sum_{k=1}^n X_k(\omega) \sim (1+\lambda)n$, hence $\sum_{n\geq 1} \frac 1{Y_n(\omega)} = +\infty$.

Therefore $\sum_{n\geq 1} \frac 1{Y_n}$ diverges almost surely.


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