Proof: By Induction
(related to Proposition: Ratio Test)
 By hypothesis, \(\sum_{n=0}^\infty a_n\) is an infinite series, and \(a_n\neq 0\) for all \(n \ge N\), where \(N\in\mathbb N\) is some index (i.e. all but the first \(N\) sequence members \(a_n\) must be different from \(0\)), and there exists a fixed^{1} positive number \(q\) with \(0 < q < 1\) such that \(\left\frac{a_{n+1}}{a_n}\right\le q\) for all \(n \ge N.\)
 We observe that changing a finite number of the sequence members does not change its convergence behavior.
 Therefore, by changing the first \(N\) sequence members \(a_n\), we can assume that \(a_n\neq 0\) for all \(n \in\mathbb N\) (i.e. all sequence members \(a_n\) are different from \(0\)), and there exists a real number \(q\) with \(0 < q < 1\) such that \(\left\frac{a_{n+1}}{a_n}\right\le q\) for all \(n \in \mathbb N.\)
 It follows from the proving principle by induction that
 Base case: $a_1\le a_0 q^0=a_0\cdot 1,$ and that
 Induction step: given \(a_n\le a_0q^{n1},\), we have $a_{n+1}\le a_n\cdot q\le a_0q^{n}.$
 Due to the convergence of the infinite geometric series $ \sum_{n=0}^\infty a_0q^{n}=\frac{a_0}{1q}$ is convergent.
 It follows from the direct comparison test for absolutely convergent series that \(\sum_{n=0}^\infty a_n\) is absolutely convergent.
 On the other hand, if $q\ge 1$ then for some index $N\in\mathbb N$ we have $\left\frac{a_{n+1}}{a_n}\right\ge 1$ which is equivalent to $a_{n+1}\ge a_n$ for all $n\ge N.$
 Therefore, the sequence $(a_n)_{n\in\mathbb N}$ is monotonically increasing and therefore it cannot converge to $0.$
 By contraposition to convergent series implies convergent sequence, the series $\sum_{n=0}^\infty a_n$ is divergent.
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References
Bibliography
 Forster Otto: "Analysis 1, Differential und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983
Footnotes