# Absolute convergence a criterion for unconditional convergence.

I'm working through Dr. Pete Clark's convergence notes here: http://math.uga.edu/~pete/convergence.pdf

and I've been thinking about Exercise 3.2.2 (b),

The question states that for the set $S = \mathbb{Z}^{+}$, the series $\sum_{i\in S}x_{i}$ converges unconditionally if and only if it converges absolutely, i.e, $\sum_{i\in S}|x_{i}| < \infty$.

This seems to contradict the well-known fact:

$\sum_{n=1}^{\infty}\frac{1}{n} = \infty$ and $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n} < \infty$.

If I take $x_{n} = \frac{(-1)^{n}}{n}$, then $|x_{n}| = \frac{1}{n}$, and it seems to violate Exercise 3.2.2(b) on page 11.

Am I missing the point behind unconditional convergence? I made the jump that unconditional convergence (as defined in the link on the top of page 11) reduces to the ordinary convergence of series when the underlying set $S$ takes the place of the positive integers. Is this incorrect?

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It would have been useful if you included what "Exercise 3.2.2(b)" states, to make the question self-contained.... –  Arturo Magidin May 9 '12 at 1:28
Sorry about that, I edited the question to reflect this change. –  Kyle Schlitt May 9 '12 at 1:34

You are misinterpreting 3.2.2(b).

In fact, $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n}$ does not converge absolutely, as you note; hence this series also does not converge unconditionally.

Indeed, for the series to converge unconditionally to $a$, it would have to be the case that for every $\epsilon\gt 0$ there is a finite set $J(\epsilon)$ such that for all finite subsets $J$ that contain $J(\epsilon)$ we have that $|a-\sum_{j\in J} a_j|\lt\epsilon$. But in this series, the positive terms diverge (as do the infinite terms). So for every $\epsilon\gt 0$ there is a finite set of even indices for which the sum is as large as we want, and so we can always find finite subsets that are arbitrarily far from any particular real number $a$.

You are also misunderstanding unconditional convergence. It does not reduce to ordinary convergence even when the index set is the positive integers. What makes you think so? Unconditional convergence in the case of sequences essentially tells you that you that given a "degree of tolerance" ($\epsilon$), there is a finite set of terms that account for "most" of the summation, in the sense that the sum of those finitely many terms is within the tolerance of the limit, and no finite number of the remaining terms will get you out of that "tolerance zone." This is different from regular convergence, in which you can only omit a "tail" of the sequence. The example of the alternating harmonic series highlights why this is very different: for any $N\gt 0$, there is always a way to pick finitely many terms "beyond the $N$th term" that add up to a lot, because the tail will take care of this by making the total contribution of the entire "tail" negligible. That is, the entire tail cannot contribute much, but in unconditional convergence no finite subset of the tail is allowed to contribute much. That's a much stronger condition than regular convergence (which makes sense given that it is equivalent to absolute convergence, which is likewise a much stronger condition than regular convergence).

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Thanks for this! –  Kyle Schlitt May 9 '12 at 1:44

Yes, "unconditional convergence (as defined in the link on the top of page 11) reduces to the ordinary convergence of series when the underlying set $S$ takes the place of the positive integers" is incorrect. The series $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}$ converges in the usual sense, but $\sum_{n \in \mathbb{N}} \frac{(-1)^n}{n}$ does not converge unconditionally.

An informal (but only slightly informal) way of making the distinction between convergence in the usual sense, and unconditional convergence, is that convergence in the usual sense only pays attention to the sequence of finite subsets $\{1\}, \{1,2\}, \{1,2,3\}, \{1,2,3,4\},\dots$ of $\mathbb{N}$, whereas unconditional convergence concerns itself with what happens with the set of all finite subsets of $\mathbb{N}$.

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Thanks very much. This went to the very root of my misunderstanding. –  Kyle Schlitt May 9 '12 at 1:38