Problem on Convergence of random series

Question

Suppose that $\{X_n\}$ is an independent sequence and $E[X_n]=0$. If $\sum \operatorname{Var}[X_n] < \infty$, then $\sum X_n$ converges with probability $1$. Is independence necessary condition here ? I am thinking of a counterexample. The intuition behind the other assumptions is clear.

Answer 1

It is. Let $X_1$ be $1$ or $-1$ with probabilities $1/2$ and let $X_n := \frac{1}{n} X_1$. series of variances is convergent, while $\sum X_n$ doesn't.

Answer 2

The condition can be relaxed, but not removed. Take for instance the following example:

$$X_n = \frac{Z}{\sqrt{n}},$$

where $Z\sim N(0,1)$. Then $E(X_n)=0$ and

$$\sum_{n=1}^\infty V(X_n) = 1,$$

but

$$\sum_{n=1}^\infty X_n= \sqrt{n} Z$$

diverges.