An interesting problem about almost sure convergence Assume that $X_1,X_2,\ldots$ are independent random variables (not necessarily of the same distribution). Assume that that $Var[X_n]>0$ for all $n$. Assume also that
$$\sum_{n=0}^\infty \frac{Var[X_n]}{n^2}<\infty,$$
that
$$\frac{1}{n}\sum_{i=1}^n(X_i-E[X_i])\to 0 \textrm{ almost surely as $n\to\infty$},$$
that
$$E[X_n]>0 \textrm{ for all $n$},$$
and that
$$\liminf_{n\to\infty} E[X_n] > 0.$$
How can we prove that $$\sum_{i=1}^n X_i \to \infty\text{ almost surely as $n\to\infty$?}$$
This seems to intuitively make sense, but a formal proof escapes me. Also, what can we say if $E[X_n] = 0$ for all $n$ and $\lim_{n\to\infty} E[X_n] = 0$?

If $\liminf E[X_n] >0$ then there exist a constant $a>0$ such that $\sum_{i=1}^{n}E[X_n]>an $ for sufficiently large $n$.
Now, reasoning with fixed $\omega$, if $\sum X_i < \infty$, then for $n$ large enough 
$$ \frac{1}{n}\sum_{i=1}^n X_i -E[X_i] < \frac{\sum_{i=1}^{\infty}X_i}{n}-a $$
where the right hand side is stricly negative for large $n$. This implies that $\frac{1}{n}\sum_{i=1}^n X_i -E[X_i]$ doesn't tend to zero.
We conclude that $\mathbb{P}(\frac{1}{n}\sum_{i=1}^n X_i -E[X_i]\rightarrow 0)=1$ implies $\mathbb{P}(\sum X_i<\infty)=0$
