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 November 6th, 2019, 10:19 AM #1 Newbie   Joined: Jan 2014 Posts: 20 Thanks: 0 absolutely convergent series Hello. I want to show that, for an absolutely convergent series $\displaystyle \sum_{n=1}^{\infty}a_n$, we have $\displaystyle \left|\sum_{n=1}^{\infty}a_n\right|\leq\sum_{n=1}^ {\infty}|a_n|$. Let $\displaystyle M$ be an positive integer. I begin with the triangle inequality $\displaystyle \left|\sum_{n=1}^{M}a_n\right|\leq\sum_{n=1}^{M}|a _n|$ and taking limit of both sides as $\displaystyle M\to\infty$. How to show that $\displaystyle \lim_{M\to\infty}\left|\sum_{n=1}^{M}a_n\right| = \left|\sum_{n=1}^{\infty}a_n\right|$? Thank you. Last edited by skipjack; November 6th, 2019 at 04:32 PM.
 November 6th, 2019, 10:49 AM #2 Senior Member   Joined: Dec 2015 From: Earth Posts: 823 Thanks: 113 Math Focus: Elementary Math $\displaystyle |a_n - L| < \epsilon \rightarrow 0$. $\displaystyle |x+y|\leq |x|+y \leq |x|+|y|$. For n-variables : $\displaystyle |x_1 + ... +x_n| \leq |x_1| + |x_2 + ... +x_n |\leq ... \leq |x_1 | +...+|x_n |$. Thanks from topsquark and woo Last edited by idontknow; November 6th, 2019 at 10:57 AM.
November 6th, 2019, 11:53 AM   #3
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Quote:
 Originally Posted by woo How to show that $\displaystyle \lim_{M\to\infty}\left|\sum_{n=1}^{M}a_n\right| = \left|\sum_{n=1}^{\infty}a_n\right|$? Thank you.
Absolute value is a continuous function which means it commutes with limits. Specifically, this implies the following equality:
$\lim_{M\to\infty}\left|\sum_{n=1}^{M}a_n\right| = \left|\lim_{M\to\infty}\sum_{n=1}^{M}a_n \right|$

Last edited by skipjack; November 6th, 2019 at 04:26 PM.

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