Proof
(related to Proposition: Direct Comparison Test For Absolutely Convergent Series)
 Let $\sum_{k=0}^\infty x_k$ and $\sum_{k=0}^\infty y_k$ be real infinite series.
 By hypothesis, let $\sum_{k=0}^\infty y_k$ be convergent.
 By the Cauchy criterion, for every \(\epsilon > 0\) there is an index \(N(\epsilon)\in\mathbb N\) such that \[\left \sum_{k=m}^n y_k \right < \epsilon\quad\quad \text{for all}\quad n\ge m\ge N(\epsilon).\]
 Since, by hypothesis, $x_k\le y_k$ for all $k\in\mathbb N,$ we have $$\left\sum_{k=m}^n x_k\right \le \left\sum_{k=m}^n y_k\right < \epsilon\quad\quad \text{for all}\quad n\ge m\ge N(\epsilon).$$
 Applying the Cauchy criterion once again, we get that the \(\sum_{k=0}^\infty x_k\) is convergent.
 By definition, \(\sum_{k=0}^\infty x_k\) is absolutely convergent.
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983