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I'm self-studying from the book Understanding Analysis by Stephen Abbott and I have a question about the proof of the Absolute Convergence Test (theorem 2.7.6 on page 65).

The author states that since the series $\sum_{n=1}^\infty |a_n|$ converges, we know that, given $\epsilon >0$, there exists an $N \in \mathbb{N}$ such that: \begin{equation} |a_{m+1}| + |a_{m+2}| + \cdots + |a_n| < \epsilon \end{equation} for all $n > m \geq N$.

My question is: why this is true? I don't think the author has proved this anywhere in the book.

Intuitively, I can sort of understand the above statement only if the "tail" of the series becomes increasingly small (which naively seems to a property of the convergent series, but I'm not sure if this is always true). But this all seems very "handwavy". Can anybody shed some light on my confusion?

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    $\begingroup$ The sequence of partial sums of $\sum_{n=1}^\infty|a_n|$ converges and is, thus, .... $\endgroup$
    – David Mitra
    Aug 20, 2014 at 14:31
  • $\begingroup$ @DavidMitra a Cauchy sequence? I'm not sure how this is going to help me though. $\endgroup$
    – Hunter
    Aug 20, 2014 at 14:37
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    $\begingroup$ Yes. What is $S_{n}-S_m$? $\endgroup$
    – David Mitra
    Aug 20, 2014 at 14:38
  • $\begingroup$ @DavidMitra $|s_n - s_m| = |a_{m+1} + a_{m+2} + \cdots + a_n | < \epsilon$. But this still doesn't tell me that $|a_{m+1}| + |a_{m+2}| + \cdots + |a_n | < \epsilon $, right? $\endgroup$
    – Hunter
    Aug 20, 2014 at 14:41
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    $\begingroup$ You're looking at the wrong series. Look at the partial sums for $\sum_{n=1}^\infty |a_n|$. You're assuming this converges. $S_n-S_m= |a_{m+1}|+\cdots+|a_n|$. $\endgroup$
    – David Mitra
    Aug 20, 2014 at 14:43

1 Answer 1

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You're assuming the series $\sum\limits_{n=1}^\infty |a_n|$ converges. This, by definition means that the sequence of partial sums, $(S_m)$, given by $S_m=\sum\limits_{n=1}^m |a_n|$, converges.

If you look closely, you should be able to see that your condition is just saying $(S_m)$ is a Cauchy sequence (which it is, of course).

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