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 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 \(|a_n|\le |a_0|q^{n-1},\), we have $|a_{n+1}|\le |a_n|\cdot q\le |a_0|q^{n}.$
- Due to the convergence of the infinite geometric series $ \sum_{n=0}^\infty |a_0|q^{n}=\frac{|a_0|}{1-q}$ 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
