# Convergence of a random harmonic series with Poisson gaps

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.)

• What is $Y_1^{-1}$ when $X_1=0$ (with probability $e^{-\lambda}$)? Apr 21, 2021 at 22:00
• 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 Apr 21, 2021 at 22:08
• Good point. The modification you propose seems reasonable and I'll adopt it. Apr 21, 2021 at 22:17
• @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.) Apr 21, 2021 at 22:22
• 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 Apr 21, 2021 at 22:38

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.