For arbitrary $\{X_n\}$ if $\sum_{n=1}^{\infty}{ \mathbb{E}(|{X_n}|)} <\infty$ then $\sum_{n=1}^{\infty}{ {X_n}} $ converges absolutely $a.e$ For arbitrary  $\{X_n\}$ if
$$\sum_{n=1}^{\infty}{ \mathbb{E}(|{X_n}|)} <\infty$$
then $\sum_{n=1}^{\infty}{ {X_n}} $ converges absolutely $a.e$
MY SOLUTION: For Three series of Kolmogorov,
Series 1 We must verify that $\sum_{n\ge 1} \mathbb{E}(|X_n| > 1) < +\infty$. Observe
\begin{eqnarray}
\sum_{n\ge 1} \mathbb{P}(|X_n| > 1) &\le& \sum_{n\ge 1} \mathbb{E} (|X_n|)  < \infty.\\
\end{eqnarray}
thus series 1 ok.
Series 2 We need to show $\sum_{n\ge 1} \mathbb{E}(|X_n| 1_{|X_n|\le 1})$ converges. by the hypothesis we have.
$$
\ \ \ \ \ \ \sum_{n\ge 1} \mathbb{E}(|X_n| 1_{|X_n|\le 1}) \le  \sum_{n\le 1}^{}{ \mathbb{E}(|{X_n}|)} <\infty.
$$
thus series 2 ok.
Series 3 We must show $\sum_{n\ge 1} \mathrm{Var}(|X_n| 1_{|X_n|\le 1}) < \infty$. Given $(*)$ above, this is easy:
\begin{eqnarray}
(*)\,\,\,\,\,\,\,\,\,\,\sum_{n\ge 1} \mathrm{Var}(|X_n| 1_{|X_n|\le 1}) &\le& \sum_{n\ge 1} \mathbb{E}(|X_n|^2 1_{|X_n|\le 1}) < \infty.
\end{eqnarray}
$(*)\,\,$ why this$\sum_{n\ge 1} \mathbb{E}(|X_n|^2 1_{|X_n|\le 1})$ series converges?
I do not know, thanks for your help.

$$\sum_{n=1}^{\infty}{ \mathbb{E}(|{X_n}|)} <\infty$$
$$\to \mathbb{E} \left[\sum_{n=1}^{\infty}{ |{X_n}|} \right] <\infty \tag{1}$$
$$\to \sum_{n=1}^{\infty}{ |{X_n}|} <\infty \ \text{a.e./a.s.} \ \tag{2}$$
$(1)$
$$E[\sum_n X_n] = \sum_n E[X_n] \ \text{if} \ X_n \ge 0$$
$(2)$
$$X = \infty \ \text{a.s./a.e.} \ \to E[X] = \infty$$
