absolute convergence implies unconditional convergence

Question

Why is it true that absolute convergence implies unconditional convergence ? This is interesting, because the sequences (corresponding to each rearrangement) of partial sums are different, still they converge to the same value if the series is absolutely convergent. I want to know the intuitive reason.

Answer 1

In my opinion, the intuitive argument should run as follows:

A series is absolutely convergent iff and only if the sequence of the partial sums of the absolute values of its terms (which is always monotonic non-decreasing) is bounded above.

Now consider any re-arrangement of this series of absolute values of the terms of the original series.

If we form a re-arrangement of this series (of absolute values) and consider any given term of its sequence of partial sums, we would find that that term would be less than or equal to some term of the (bounded above) sequence of partial sums of the absolute values of the terms of the original (absolutely convergent) series.

Thus the sequence of the partial sums of this re-arrangement, which is again non-decreasing, is also bounded above.

Hence the re-arrangement is also (absolutely) convergent.

Answer 2

Unconditional convergence means the convergence to a common limit of every sequence $(\sigma_k)$ where $\sigma_k=\sum\limits_{n\in N(k)}x_n$ as long as the finite sets $N(k)$ are such that $N(k)\subseteq N(k+1)$ for every $k$ and $\bigcup\limits_kN(k)=\mathbb N$. In particular, for every $n$ such that $N(k)\supseteq\{1,2,\ldots,n\}$, $\sigma_k$ is between $s_n-r_n$ and $s_n+r_n$, where $s_n=\sum\limits_{i\leqslant n}x_i$ and $r_n=\sum\limits_{i\geqslant n+1}|x_i|$. Thus:

• Absolute convergence is the condition that $r_n\to0$ when $n\to\infty$.
• The sequence of sets $(N(k))$ is exhausting in the sense that one can choose $n_k$ such that $N(k)\supseteq\{1,2,\ldots,n_k\}$ and $n_k\to\infty$ when $k\to\infty$.
• One has $s_n\to s$ when $n\to\infty$, where $s=\sum\limits_{i\geqslant 1}x_i$.
• And $\sigma_k-s_{n_k}\to0$ when $k\to\infty$ because $n_k\to\infty$ guarantees that $r_{n_k}\to0$.

Hence $\sigma_k\to s$ when $k\to\infty$, for every exhausting sequence of sets $(N(k))$.

Answer 3

One proof of absolute convergence happening if and only if we have unconditional convergence works by showing that absolute convergence happens if and if both the series of positive terms and the series of negative terms converge. The intuitive description I guess would be that conditional convergence arises when (and only when) you get situations where the series of positive and negative terms both diverge, but summed together they converge.

Then the interesting part of this is that when we have a conditionally convergent series, how much the divergence of each part of the series "cancels out" the divergence of the other one to sum to a finite number depends only on the arrangement, rather than what the actual numbers are. It's not too shocking though considering the sequence of terms always converges to 0.