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: Let $\sum \limits_{n=1}^{\infty}|a_n|<\infty. $
For any $n\in \mathbb{N}$ we have triangle inequality: $$|a_1+a_2+\cdots+a_n|\leqslant |a_1|+|a_2|+\cdots+|a_n|\leq \sum \limits_{n=1}^{\infty}|a_n| .$$
This implies $\left|\sum \limits_{j=1}^{n}a_j\right|\leqslant \sum \limits_{n=1}^{\infty}|a_n|$ for each $n\in\mathbb{N}.$ Letting $n\to\infty$ in the last inequality and using continuity of modulus function, we get
$$\left|\sum \limits_{n=1}^{\infty}a_n\right|\leqslant \sum \limits_{n=1}^{\infty}|a_n|.$$
If $\sum \limits_{n=1}^{\infty}|a_n|=\infty$ then inequality is true.
A: 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.
A: I think in this case the easiest way is to show it via contradiction. Let's assume assume that $\sum\limits_{n=1}^{\infty}|a_n|$ and $\sum\limits_{n=1}^{\infty}a_n$ converge.
Now, we assume $\Big|\sum\limits_{n=1}^{\infty}a_n\Big|>\sum\limits_{n=1}^{\infty}|a_n|$ and therfore we define $\epsilon:=\Big|\sum\limits_{n=1}^{\infty}a_n\Big|-\sum\limits_{n=1}^{\infty}|a_n|>0$ . Then, by definition of convergence there exists an index $m_0$ such that for all $m>m_0$ we have $\Big|\sum\limits_{n=1}^{\infty}a_n-\sum\limits_{n=1}^{m}a_n\Big|<\frac{\epsilon}{2}\implies \Big|\sum\limits_{n=1}^{m}a_n\Big|>\sum\limits_{n=1}^{\infty}|a_n|\geq \sum\limits_{n=1}^{m}|a_n|$, which contradicts triangle inequality. Note that if $\sum\limits_{n=1}^{\infty}|a_n|=\infty$ then we get a contradiction immediately.
