## Calculus: Early Transcendentals 8th Edition

$|\sum\limits_{i =1}^{n}a_{i}| \leq \sum\limits_{i =1}^{n}|a_{i}|$
$|\sum\limits_{i =1}^{n}a_{i}| \leq \sum\limits_{i =1}^{n}|a_{i}|$ Use the triangular inequality $|a+b|\leq |a|+|b|$ Expand the summation on both sides. $a_{1}+a_{2}+....+a_{n}=a_{1}+a_{2}+....+a_{n}$ Take the absolute values. $|a_{1}+a_{2}+....+a_{n}|=|a_{1}+a_{2}+....+a_{n}|$ Thus, $|a_{1}+a_{2}+....+a_{n}|\leq|a_{1}|+|a_{2}|+....+|a_{n}|$ (Triangular inequality $|a+b|\leq |a|+|b|$) Hence, $|\sum\limits_{i =1}^{n}a_{i}| \leq \sum\limits_{i =1}^{n}|a_{i}|$