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?

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.
