The problem is to show that the random harmonic series

$X_n:=\sum_{n=1}^{\infty}\frac{\nu_n}{n}$ with $P[\nu_n = 1] = P[\nu_n = -1] = \frac{1}{2}$


It is obvious that the harmonic series diverges and the alternating harmonic series converges. Also I am positive that the random harmonic series converges almost surely.

Say I want to prove this with Kolmogorov's 3 series theorem where I need to show convergence of the three following series.

  1. $\sum_{n=1}^{\infty}P[X_n \neq Y_n]$

  2. $\sum_{n=1}^{\infty} E[Y_n]$

  3. $\sum_{n=1}^{\infty}var[Y_n]$

where $Y_n:= X_n 1_{\{|X_n|\leq c\}}$

I already have
$E[\nu_n] = 0$
$E[X_n] = 0$
$var[\frac{\nu_n}{n}] = \frac{1}{n^2}$ and
$var[\sum_{n=1}^{\infty}\frac{\nu_n}{n}] = \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$

I need help with 1. and I am not really sure what the difference between $X_n$ and $Y_n$ is. Is it just enough to say that since $X_n \geq Y_n$ and $\sum_{n=1}^{\infty} E[X_n]$ converges, 2. converges and $\sum_{n=1}^{\infty}var[X_n]$ converges, so 3. converges?

I don't really understand this. Any help would be great.


I think you meant to say this instead: study the convergence of the random variable $\sum_{n=1}^\infty X_n$, where $X_n := \frac{\nu_n}n$. Then if you set $c = 2$, then $Y_n = X_n$ with probability 1, and all three conditions are easily seen to be satisfied.

  • $\begingroup$ Ups. Sure, that's what I meant. With that nudge it was easy. Thanks. $\endgroup$ – sebastian Jun 17 '14 at 11:08

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