Science:Math Exam Resources/Courses/MATH220/April 2005/Question 05/Solution 1

In the base case when $n=1$, we have $|a_{1}|\leq |a_{1}|$, so the statement is true.

For the inductive step, we assume that $\left|\sum _{i=1}^{n}a_{i}\right|\leq \sum _{i=1}^{n}|a_{i}|$ holds for some (fixed) $n\in \mathbb {N}$. Then we show that it also holds with $n$ replaced by $n+1$:

${\begin{aligned}\left|\sum _{i=1}^{n+1}a_{i}\right|&=\left|\sum _{i=1}^{n}a_{i}+a_{n+1}\right|\\&\leq \left|\sum _{i=1}^{n}a_{i}\right|+|a_{n+1}|\\&\leq \sum _{i=1}^{n}|a_{i}|+|a_{n+1}|\\&=\sum _{i=1}^{n+1}|a_{i}|.\end{aligned}}$

The first inequality is true by the triangle inequality and the second by the induction hypothesis. This finishes the inductive step and so the result is true for all $n\in \mathbb {N}$ by induction.