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 nvariables : $\displaystyle x_1 + ... +x_n \leq x_1 + x_2 + ... +x_n \leq ... \leq x_1  +...+x_n $. Last edited by idontknow; November 6th, 2019 at 10:57 AM. 
November 6th, 2019, 11:53 AM  #3  
Senior Member Joined: Sep 2016 From: USA Posts: 683 Thanks: 456 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
\[ \lim_{M\to\infty}\left\sum_{n=1}^{M}a_n\right = \left\lim_{M\to\infty}\sum_{n=1}^{M}a_n \right \] Last edited by skipjack; November 6th, 2019 at 04:26 PM.  

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absolutely, approach, convergent, infinity, partial, series, sum 
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