The following proposition is extracted from Topics in Banach Space Theory by Albiac and Kalton.
Proposition $8.2.2$: An operator $T:X\to X$ is absolutely summing if and only if $\sum_{n=1}^\infty \|Tx_n\|<\infty$ whenever $\sum_{n=1}^\infty x_n$ is unconditionally convergent.
In the book, the authors do not provide a proof of the above theorem and they say the proof is routine. However, I do not know how to prove it. Any hint is appreciated. If anyone knows of any reference containing a proof of the proposition, may I have it?
Definitions: Let $X$ and $Y$ be Banach spaces. $T$ is said to be absolutely summing if there is a constant $C$ such that for all choices of $(x_k)_{k=1}^n$ in $X,$ $$\sum_{k=1}^n\|Tx_k\|\leq C\sup\bigg\{ \sum_{k=1}^n|x^*(x_k)|:x^*\in X^*, \|x^*\|\leq 1 \bigg\}.$$
We say that $\sum_{n=1}^\infty x_n$ is unconditionally convergent if one of the following holds:
$(1)$ $\sum_{n=1}^\infty x_{\pi(n)}$ converges for every permutation $\pi$ of $\mathbb{N}.$
$(2)$ The series $\sum_{k=1}^\infty x_{n_k}$ converges for every increasing sequence $(n_k)_{k=1}^\infty$
$(3)$ The series $\sum_{n=1}^\infty \varepsilon_n x_n$ converges for every choice of signs $(\varepsilon_n).$
$(4)$ For every $\varepsilon>0,$ there exists $n$ such that if $F$ is a finite subset of $\{n+1,n+2,...\},$ then $$\bigg\| \sum_{j\in F} x_j \bigg\|<\varepsilon.$$