Proof
(related to Proposition: Limit Comparizon Test)
 By hypothesis, $\sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$ are infinite series with positive terms $a_n > 0$ and $b_n > 0$ for all $n\in\mathbb N$. Moreover the sequence of ratios $\left(\frac{a_n}{b_n}\right)_{n\in\mathbb N}$ converges to a positive limit $\lim_{n\to\infty }\frac{a_n}{b_n}=c > 0.$
 By the definition of convergence, there exists an index $N$ such that for all $n\ge N$ we have $$0 < \alpha:=\frac c2 < \left(\frac{a_n}{b_n}\right) < \frac {3c}2=:\beta.$$
 Thus, for all $n\ge N$ we have $0 < \alpha b_n < a_n < \beta b_n.$
 It follows from the direct comparison test for absolutely convergent series that the sequences $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ are either both absolutely convergent or both or both divergent.
 If the limit is zero ($c=0$), then there is some index $N$ such that for all $n\ge N$ we have $0 < \left(\frac{a_n}{b_n}\right) \le 1,$ or $0 < a_n <\le b_n.$
 The direct comparison test for absolutely convergent series shows that if $(a_n)_{n\in\mathbb N}$ is absolutely convergent, if $(b_n)_{n\in\mathbb N}$ is convergent.
 If $(b_n)_{n\in\mathbb N}$ is divergent, no conclusion whatsoever on the convergence behavior of (a_n)_{n\in\mathbb N}$ can be made.
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References
Bibliography
 Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition