# Proof: By Induction

Note that changing a finite number of the sequence members of any infinite complex series $$\sum_{k=0}^\infty a_k$$ does not change its convergence behavior. Therefore, by changing the first $$K$$ (if any) sequence members $$a_k$$, we can assume that

1. $$a_k\neq 0$$ for all $$k \in\mathbb N$$ (i.e. all sequence members $$a_k$$ are different from complex zero $$0$$),
2. there exists a real number $$\theta$$ with $$0 < \theta < 1$$ such that $$\left|\frac{a_{k+1}}{a_k}\right|\le \theta$$ for all $$k \in \mathbb N.$$

It follows from the proving principle by induction that (base case) $|a_1|\le |a_0|\theta^0=|a_0|\cdot 1,$ and that (induction step), given $$|a_k|\le |a_0|\theta^{k-1},$$ $|a_{k+1}|\le |a_k|\cdot \theta\le |a_0|\theta^{k}.$

From this result, it follows that the series $$\sum_{k=0}^\infty |a_0|\theta^{k}$$ is a majorant of the series $$\sum_{k=0}^\infty a_k$$. Due to the convergence of the infinite geometric series. $\sum_{k=0}^\infty |a_0|\theta^{k}=|a_0|\sum_{k=0}^\infty \theta^{k}=\frac{|a_0|}{1-\theta}$ it follows from the majorant criterion for complex series that $$\sum_{k=0}^\infty a_k$$ is an absolutely convergent complex series.

Github: ### References

#### Bibliography

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