Triangle inequality for infinite number of terms We can prove that for any $n\in \mathbb{N}$ we have triangle inequality: $$|x_1+x_2+\cdots+x_n|\leqslant |x_1|+|x_2|+\cdots+|x_n|.$$
How to prove it for series i.e. $$\left|\sum \limits_{n=1}^{\infty}a_n\right|\leqslant \sum \limits_{n=1}^{\infty}|a_n|.$$
Can anyone help to me with this?

A slightly different proof for the case $\sum_{n=1}^\infty |a_n|< \infty $ (i.e., $\sum_{n=1}^\infty a_n$ is absolutely convergent):
First, we show that $\lim_{N \to \infty} |\sum_{n=1}^N a_n| = | \sum_{n=1}^\infty a_n|$.
Denote the $N$th partial sum by $s_N := \sum_{n=1}^N a_n$.
The absolute convergence of $\sum_{n=1}^\infty a_n$ implies ordinary convergence, so there exists $L \in \mathbb{R}$, s.t. $L = \sum_{n=1}^\infty a_n := \lim_{N\to \infty} s_N $. Then by this exercise, the absolute value of $s_N$ also converges to the absolute value of $L$, i.e.,  $\lim_{N\to \infty} |s_N| = |L| = | \sum_{n=1}^\infty a_n|$ .
Finally, $\forall N, |s_N| = |\sum_{n=1}^N a_n|\leq \sum_{n=1}^N |a_n| $ (by finite triangle inequality). Taking $\lim_{N\to \infty}$ on both sides gives the desired result.
