Divergence of random series So, suppose that $X_n$ is a sequence of independent identically distributed random variables with Bernoulli distribution with parameter $p$. Now, consider the random series $$\sum_{n=1}^{\infty} \frac{X_n}{n}$$
Well, I need to show that if $p>\frac{1}{2}$ then the probability that this random series is divergent is 1.
Try: Well, I think we can use a coupling argument in the case of $p=\frac{1}{2}$ but I don't know how to proceed in the given situation. Could anyone help me?
 A: The @Michele Triestino's answer is more elegant, but here is another way.
The Kolmogorov's three-series theorem states the following:
Let $Z_1,Z_2,..,$ independent random variables. Let $A>0$ and $Y_n = Z_n 1_{\{|Z_n| \le A\}}$. So, $\sum_{n=1}^{\infty}Z_n$ converges if, and only if,


*

*$\sum_{n=1}^{\infty} P(|Z_n| >A) < \infty$

*$\sum_{n=1}^{\infty}EY_n < \infty$

*$\sum_{n=1}^{\infty}var(Y_n) < \infty$


Now, put $Z_n = \frac{X_n}{n}$. So, $Z_n$'s are independent and, for large enough $n$, $Z_n = Y_n$. But $\sum_{n=1}^{\infty}EY_n = \sum_{n=1}^{\infty}\frac{p}{n}$ which diverges.
In fact, this means that the random series converges if, and only if, $EX_n = 0$
A: The case $p=\frac12$ and $X_n=\pm 1$ is treated in Lawler's book "Introduction to Stochastic Process", along a list of examples in the different paragraphs of Section 5.
I will resume the main points:


*

*The random variables $Y_n=\frac{X_n}{n}$ are independent and their mean is zero. This implies that the partial sum $M_N=\sum_{n=1}^NY_n$ is a martingale, and actually uniformly integrable as $\sum_{n=1}^\infty\frac{1}{n^2}<\infty$.

*The uniform integrability of the martingale implies that $M_N$ converges to a random variable $M_\infty$ with expectation $E[M_\infty]=E[M_1]=0$. So the probability that the random sum is infinite is zero.
Now, let us consider independent random variables $X_n$ of mean $E[X_n]=\mu$. Then I define the random variables $Y_n=\frac{X_n}{n}-\frac{E[X_n]}{n}$ and observe that the random sum $M_N=\sum_{n=0}^NY_n$ is a uniformly integrable martingale, so it converges to some random variable $M_\infty$.
This means that the random infinite sum
$$\sum_{n=1}^\infty\left(\frac{X_n}{n}-\frac{\mu}{n}\right)$$
is finite with probability $1$.
This implies that the random series is convergent if and only if $E[X_n]=\mu=0$.
