Proof of equivalent condition for $\sum_{x\in X}f(x)$ being absolutely convergent (Lemma 8.2.3 in Tao's Analysis 1) 
Lemma  8.2.3 Let $X$ be a countable set, and let $f:X \rightarrow\mathbb R$ be a function. Then the series $ \sum_{x \in X}  f(x)$ is absolutely convergent if and only if
  $$
\sup \left\{ \sum_{x \in A} |f(x)|: A \subseteq X, A\text{ finite}  \right\} < \infty 
$$.

First, I have no idea how to prove it formally. As far as idea about the theorem is concerned, I can sense that if $\sum_{x \in X}  |f(x)| \leq  L$ since it is convergent (say to $L'$ which is $\leq L$). So, any summation over any finite set will be less than $\infty$. But, what about converse of the theorem and how to prove these result.
for converse part: 
As a counterexample, this result does not seem to hold for when $f(n)= \frac{1}{n}$ as Supremum of the above set seems to be finite for any finite set but $\sum_{x \in X} f(x)$ is divergent.
 A: Let $g: \Bbb N \to X$ be a bijection. If $\sum_{x \in X} f(x)$ is absolutely convergent, then $\sum_{n \in \Bbb N} |f(g(n))|$ is convergent, hence:
$$\infty > \sum_{n=0}^{\infty} |f(g(n))| = : L.$$ 
Given $A \subset X$ a finite set, we have:
$$\sum_{x \in A} |f(x)| = \sum_{n \in g^{-1}(A)} |f(g(n))| \le \sum_{n \in \Bbb N} |f(g(n))| = L$$
Hence
$$\sup\left\{ \sum_{x \in A} |f(x)|, A \subseteq X \text{ finite } \right\} \le L < \infty$$
Edit
Suppose that:
$$\sup\left\{ \sum_{x \in A} |f(x)|, A \subseteq X \text{ finite } \right\} =: L < \infty$$
Let $n \in \Bbb N$. Then, $g(\{1,\ldots, n\}) =:A$ is a finite subset of $X$. Hence,
$$\sum_{x \in A} |f(x)| \le L$$
But:
$$\sum_{x \in A} |f(x)| = \sum_{x\in g(\{1,\ldots, n\}) } |f(x)| = \sum_{k \in \{1,\ldots, n\}} |f(g(k))| = \sum_{k=1}^n |f(g(k))|.$$
Hence
$$\sum_{k=1}^n |f(g(k))| \le L.$$
and this is true for all $n \in \Bbb N$. 
This shows that the series $\sum_{n \in \Bbb N} |f(g(n))|$ is convergent. Therefore, $\sum_{x \in X} f(x)$ is absolutely convergent.
