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Prove the triangle inequality for series, that is if $\sum x_n$ converges absolutely then $|\sum_{n=1}^{\infty}x_n| \ge \sum_{n=1}^{\infty}|x_n| $.

My attempy: Since $\sum x_n$ converges absolutely, it is cauchy series. Then, $\forall \varepsilon>0,$ there exists $M\in Z^+$ such that $|\sum_{j=n+1}^{k}|x_n||<\varepsilon$ for $n\ge M$ and $k>n$. Then,

$|\sum_{j=n+1}^{k}|x_n||=\sum_{j=n+1}^{k}|x_n| \ge |\sum_{j=n+1}^{k}x_n|$ by triangle inequality.

I think it is wrong since I use triangle inequality, which is required to be proven, and I also don't know how to change $\sum_{j=n+1}^{k} \rightarrow \sum_{j=1}^{\infty}$.

Could you give some help?

Thank you in advance.

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  • $\begingroup$ Isn't the inequality the other way around? $\endgroup$ – Bill O'Haran Apr 17 '18 at 10:29
  • $\begingroup$ Hint: if you have two convergent sequences $a_n$ and $b_n$ and you know $a_n\leq b_n$ for every $n$, what can you say in the limit? $\endgroup$ – Uskebasi Apr 17 '18 at 10:31
  • $\begingroup$ this $|\sum_{n=1}^{\infty}x_n| \ge \sum_{n=1}^{\infty}|x_n|$ is wrong. The triangle inequality is this: $$|\sum_{n=1}^{\infty}x_n| \le \sum_{n=1}^{\infty}|x_n|$$ Use the algebra of limits, that is, if $\lim a_n=a$ and $\lim b_n=b$ and $a_n\le b_n$ for all $n\in\Bbb N$ then $a\le b$ $\endgroup$ – Masacroso Apr 17 '18 at 10:33
  • $\begingroup$ Actually, $b_n$ convergent is enough $\endgroup$ – Uskebasi Apr 17 '18 at 10:33
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we have

$|\sum_{j=1}^{n}x_j|\leq \sum_{j=1}^{n}|x_j|$ by the triangle inequality wich is true for any $n \in\mathbb{N}$ so by taking $ n \rightarrow +\infty $ you can deduce $|\sum_{n=1}^{\infty}x_n| \leq \sum_{n=1}^{\infty}|x_n|$

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  • $\begingroup$ The question asks to prove triangle inequality, so I think we cannot use that property. $\endgroup$ – user1230403 Apr 17 '18 at 10:39
  • $\begingroup$ yes we can because i use this inequality for a finite number of terms , after the inequality is true for any $n$. so we can use the limite $\endgroup$ – Bernstein Apr 17 '18 at 10:41
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    $\begingroup$ The question asks to prove triangle inequality for series. You can use the standard triangle inequality $\endgroup$ – Uskebasi Apr 17 '18 at 10:41
  • $\begingroup$ you are using that $|\sum x_j| \le \sum |x_j|$ at first. Isn't it what we want to prove? $\endgroup$ – user1230403 Apr 17 '18 at 10:49
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    $\begingroup$ i use this $|\sum_{j=1}^{n}x_j|\leq \sum_{j=1}^{n}|x_j|$ and this is true by the classical triangle inequality $\endgroup$ – Bernstein Apr 17 '18 at 10:51

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