Suppose $(X,S,\mu)$ is a measure space. Suppose $f_n$ is integrable for each $n$.
Suppose $\sum_{n=1}^{\infty}\int |f_n|d\mu<\infty$.
Then, $\lim_{N\rightarrow \infty} \sum_{n=1}^N \int |f_n|d\mu=\lim_N \int \sum_{n=1}^N |f_n| d\mu<\infty$. Hence, setting $g_N= \sum_{n=1}^N|f_n|$ for each $N$, we obtain a non-negative increasing sequence:
$g_1\leq g_2\leq g_3.....$, where each $g_N$ is integrable.
So by Fatou's lemma,
$\int \lim_n \inf \sum_{n=1}^N|f_n| d\mu\leq lim_N \inf \int g_N d\mu =\lim \inf \int \sum_{n=1}^N |f_n| d\mu=\sum_{n=1}^{\infty}\int |f_n|d\mu$.
I'm not sure sure what else can I conclude, may I have some help?
Assume $(f_n)$ is a sequence of non-negative integrable functions with $\sum_{n=1}^{\infty} \int f_n < \infty$. You can apply the MCT to $g_N = \sum_{n=1}^{N} f_n $ with $g = \sum_{n=1}^{\infty} f_n$ to see that this integrable and that $\sum_{n=1}^{\infty} f_n$ converges to a finite a.e. function.
For a general function if $\sum_{n=1}^{\infty} \int | f_n| < \infty$ then you can use the DCT to demonstrate that $\sum_{n=1}^{\infty} f_n $ converges (note if this condition does not hold then the series does not converge). Is it clear now?