We provide a proof by induction for natural numbers $n\ge 1$ and absolute value of real numbers”:https://www.bookofproofs.org/branches/absolute-value-of-real-numbers-modulus/ (respectively absolute value of complex numbers).

- Base case: $n=1$

$$\left|\sum_{k=1}^1 a_k\right|=|a_1|\le|a_1|=\sum_{k=1}^1|a_k|.$$ - Induction step $n\to n+1$
- Assume, $$\left|\sum_{k=1}^n a_k\right|\le\sum_{k=1}^n |a_k|$$ holds for some $n\ge 1.$
- Then, by the triangle inequality $$\begin{align}\left|\sum_{k=1}^{n+1} a_k\right|&\le \left|\sum_{k=1}^{n} a_k\right|+|a_{n+1}|\nonumber\\

&\le\sum_{k=1}^{n} |a_k|+|a_{n+1}|\quad\text{(by assumption)}\nonumber\\

&= \sum_{k=1}^{n+1} |a_k|.\nonumber\end{align}$$

q.e.d

| | | | created: 2020-06-26 16:53:47 | modified: 2020-06-26 16:54:00 | by: | references: [8311]

[8311] **Modler, F.; Kreh, M.**: “Tutorium Analysis 1 und Lineare Algebra 1”, Springer Spektrum, 2018, 4. Auflage