# 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?

• where is it you get stuck? – Zelos Malum Dec 15 '15 at 14:27
• You need more conditions, e.g. the inequality doesn't make sense if either $\sum_{n=1}^{\infty}a_n$ or $\sum_{n=1}^{\infty}|a_n|$ diverges. – user236182 Dec 15 '15 at 14:28
• @user236182 The divergence of $\sum \lvert a_n\rvert$ is harmless, we have an obvious (and correct) interpretation then, provided $\sum a_n$ converges. – Daniel Fischer Dec 15 '15 at 14:30
• Write each series as a limit of partial sums. If the inequality holds for each partial sum, it must hold in the limit. – kccu Dec 15 '15 at 14:30
• @user236182 can you give a whole claim with proof? – Raheem Najib Dec 15 '15 at 14:30

Let $\sum \limits_{n=1}^{\infty}|a_n|<\infty.$ For any $n\in \mathbb{N}$ we have triangle inequality: $$|x_1+x_2+\cdots+x_n|\leqslant |x_1|+|x_2|+\cdots+|x_n|\leq \sum \limits_{n=1}^{\infty}|a_n| .$$ 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.
Alternative proof for the case $\sum_{n=1}^\infty |a_n|< \infty$ (i.e., $\sum_{n=1}^\infty a_n$ is absolutely convergent), from first principles:
First, we need to show the sequence $(s_N = |\sum_{n=1}^N a_n|)$ converges to $| \sum_{n=1}^\infty a_n|$.
The absolute convergence of $\sum_{n=1}^\infty a_n$ implies ordinary convergence, so there exists $L \in \mathbb{R}$, s.t. $L = \sum_{n=1}^\infty a_n := \lim_{N\to \infty} \sum_{n=1}^N a_n$. Define a new sequence $(s_N)$ by $s_N = |\sum_{n=1}^N a_n|$, then $\lim_{N\to \infty} s_N=\lim_{N\to \infty} |\sum_{n=1}^N a_n | = |L|$ (see this exercise).
Finally, $\forall N, s_N = |\sum_{n=1}^N a_n|\leq \sum_{n=1}^N |a_n|$ (again by finite triangle inequality). Taking $\lim_{N\to \infty}$ on both sides gives the desired result.