# 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 fixed1 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.

Github: ### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983

#### Footnotes

1. Please note that you have to find a fixed number $q$ that is $< 1.$ In order to be able to apply the ratio test, it does not suffice to show that $\left|\frac{a_{n+1}}{a_n}\right| < 1$ for all $n > N,$ the inequality must be $\left|\frac{a_{n+1}}{a_n}\right| < q < 1.$