Proving that unconditional convergence is equivalent to absolute convergence Regarding this discussion here:
Absolute convergence a criterion for unconditional convergence.
(thank you for the great answers, by the way)
I'm still trying to do Exercise 3.2.2 (b) from these notes by Dr. Pete Clark.
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, $\sum_{i\in S}|x_{i}| < \infty$.
Thanks to the prior discussion, I get why unconditional convergence isn't simply the same thing as regular convergence in the case of $S = \mathbb{Z}^{+}$.  Unconditional convergence (to $x$) even allows for "cherry picking" of terms in the tail, rather than only being allowed to take them in order, and still reguires the resulting sums to stay within the $\epsilon$-band (of $x$).  (Again, thanks for the detailed answers to my previous question on this.)
My "best" attempt hits a brick wall:
Let $S = \mathbb{Z}^{+}$.  Write $S = S^{+}\cup S^{-}$, where
$$S^{+} := \{n\in S : x_{n} \geq 0\}$$
and
$$S^{-} := \{n\in S : x_{n} < 0\}$$
$(\Rightarrow)$ Now if I assume that $\sum_{n=1}^{\infty}|x_{n}|$ is not finite ( that is, the series is not absolutely convergent ),  I want to show that the unconditional convergence definition fails.  I want to try to construct arbitrarily large finite subsets $K$ of $S^{+}$ or $S^{-}$ such that $\sum_{i\in K}x_{i}$ is arbitrarily large, which would prove the contrapositive.  I think I can do it if one of $S^{+}$ or $S^{-}$ is finite, but this need not be the case.
Any advice or suggestions?  I feel like I hit a brick wall.
Edit:  I added the solution as an answer below for anyone who might have this same question.  Thanks all for the help.
 A: I am very flattered that you're reading my notes and taking them so seriously as to try to do the exercises!
This business of absolute / nonabsolute / conditional / unconditional convergence is really more a calculus / analysis topic than a topology topic.  Accordingly, a much more in depth treatment of this is given in $\S 14.1.2$ of my honors calculus notes.
(The notes linked to above do go on to talk about "unordered summation", which really is convergence with respect to a certain net.  In an earlier draft I had some remarks about nets, Moore-Smith convergence, and so forth: the basic papers by Moore-Smith and Tukey still appear in the bibliography.  Honestly, I feel like I regained a bit of my senses by excising this material from this "honors calculus" text.)
Let me say that thinking about it for yourself is commendable.  It's probably better for you in the long run that you first ran into this fact in a context in which the proof was not immediately provided.
A: Hint: If $\sum\limits_{n=1}^\infty |x_n|$ is infinite, then at least one of  $\sum_{S^+ } x_n$ and  $\sum_{  S^-} x_n$ must be infinite. Otherwise,   $\sum\limits_{n=1}^\infty |x_n|$ would converge, as it could be written as the difference of two convergent series (pad the positive terms of $(x_n)$ with zeroes, and the negative terms  with zeros to produce them). 
A: With the helpful hints of others, I finally put together what I think is a full solution.
Reminder:  $S = \mathbb{Z}^{+}$ and $S^{+} = S\cap (0,\infty)$ and $S^{-} = S\cap (-\infty,0)$.
$(\Rightarrow)$ I argue by contrapositive:  Suppose $\sum_{n=1}^{\infty}|x_{n}|$ is not finite.  Then 
Claim: (from the hint) we have that at least one of $\sum_{n=1}^{\infty}f^{+}(x_{n})$ and $\sum_{n=1}^{\infty}f^{-}(x_{n})$ must be infinite, where
$f^{+}(x) = \begin{cases}x & x\geq 0\\ 0 & x < 0\end{cases}$
and $f^{-}(x) = \begin{cases}x & x < 0\\ 0 & x \geq 0\end{cases}$
(I assume this is what was meant by "padding" with $0$'s)
Indeed suppose they were both finite, then we could write
$\begin{eqnarray*}
\sum_{n=1}^{\infty}|x_{n}| &=& \sum_{n=1}^{\infty}f^{+}(x_{n}) - f^{-}(x_{n})\\
 &=& \sum_{n=1}^{\infty}f^{+}(x_{n}) - \sum_{n=1}^{\infty}f^{-}(x_{n})\\
 &=& \sum_{n=1}^{\infty}f^{+}(x_{n}) - \sum_{n=1}^{\infty}f^{-}(x_{n})\\
\end{eqnarray*}$
which is the difference of two finite sums, this contradicts our original assumption and thus proves the claim.
Observation:  Using this claim, we may assume without loss of generality that $\sum_{n=1}^{\infty}f^{+}(x_{n})$ is infinite.
Then for each $m\geq 1$, define the finite set $K_{m}:=\{1, \dots, m\}\cap S^{+}$, which is finite.  Since $\sum_{n=1}^{\infty}f^{+}(x_{n})$ is infinite, the sequence
$$\sum_{n\in K_{m}}x_{n}$$ is unbounded as $m$ grows.
Now for the proof:
Now assume for a contradiction that $\sum_{n\in S}x_{n}$ converges unconditionally, say to $x\in\mathbb{R}$.  Then (taking $\epsilon = 1$ in the definition of unconditional convergence) there exists a finite subset $J\subset S$ such that whenever $K\subset S$ is finite with $J\subset K$, we have
$$|\sum_{n\in K}x_{n} - x| < 1$$
By the triangle inequality, this provides the upper bound on $\sum_{n\in K}x_{n}$:
$$|\sum_{n\in K}x_{n}|\leq 1 + |x|$$
and this bound is satisfied so long as $J\subset K\subset S$ and $K$ is finite.  Write $J^{+} = J\cap S^{+}$.
Apply this to the set $K_{m}\cup J$, for each $m\geq 1$, then we get
$\begin{eqnarray*}
|\sum_{n\in K_{m}\cup J}x_{n}| &=& |\sum_{n\in K_{m}}x_{n} + \sum_{n\in J}x_{n} - \sum_{n\in K_{m}\cap J}x_{n}|\\
&\geq& |\sum_{n\in K_{m}}x_{n} + \sum_{n\in J}x_{n}| - |\sum_{n\in K_{m}\cap J}x_{n}|\\
&\geq& |\sum_{n\in K_{m}}x_{n} + \sum_{n\in J}x_{n}| - |\sum_{n\in J^{+}}x_{n}|\\
\end{eqnarray*}$
where the last line uses $J^{+}\supset K_{m}\cap J$ and the fact that all terms "over" both set are non-negative.
This contradicts the fact that $|\sum_{n\in K_{m}\cup J}x_{n}|$ is bounded, $|\sum_{n\in J^{+}}x_{n}|$ is a constant, and  $|\sum_{n\in K_{m}}x_{n} + \sum_{n\in J}x_{n}|$ is unbounded.
$(\Leftarrow)$  Suppose $\sum_{n=1}^{\infty}|x_{n}|< \infty$.  Then by completeness of the reals, there is $x\in\mathbb{R}$ such that $\sum_{n=1}^{N}x_{n} \to x$ as $N\to\infty$.
Let $\epsilon > 0$ be given.
Choose $N_{1}\geq 1$ so that $\sum_{n=m}^{\infty}|x_{n}|\leq \frac{\epsilon}{2}$ whenever $m\geq N_{1}$.
Choose $N_{2}\geq 1$ so that $|\sum_{n=1}^{m}x_{n} - x| <\frac{\epsilon}{2}$ whenever $m\geq N_{2}$.
Set $N = max(N_{1},N_{2})$, and set $J = \{1, ... , N\}$.  Then whenever $K\subset S$ is finite such that $J\subset K$, we have
$\begin{eqnarray*}
|\sum_{n\in K}x_{n} - x| &=& |\sum_{n\in J}x_{n} - x + \sum_{n\in K\backslash J}x_{n}|\\
&\leq& |\sum_{n\in J}x_{n} - x| + |\sum_{n\in K\backslash J}x_{n}|\\
&\leq& |\sum_{n\in J}x_{n} - x| + \sum_{n\in K\backslash J}|x_{n}|\\
&\leq& \frac{\epsilon}{2} + \sum_{n = N + 1}^{\infty}|x_{n}|\\
&\leq& \epsilon
\end{eqnarray*}$
So the series converges to $x$ unconditionally.
I realize this could probably be done in a tighter fashion and is too long for me to ask anyone to check it over, but for now I'm just glad I see it.  Thanks very much for the hints, and if there are any glaring problems that someone may find in my argument please tell me I would like to fix them.
