Why is absolute convergence necessary for the Rearrangement Theorem?

Question

The following proof of the rearrangement theorem is from Bartle and Sherbert's Introduction to Real Analysis, 3rd edition. I do not see where absolute convergence was used in the proof.

9.1.5 Rearrangement Theorem Let $\sum x_n$ be an absolutely convergent series in $\mathbb{R}$. Then any rearrangement $\sum y_k$ of $\sum x_n$ converges to the same value.

Proof. Suppose that $\sum x_n$ converges to $x \in \mathbb{R}$. Thus, if $\varepsilon > 0$, let $N$ be such that if $n, q > N$ and $s_n := x_1 + \cdots + x_n$, then

$$ |x - s_n| < \varepsilon $$

and

$$ \sum_{k=N+1}^{q} |x_k| < \varepsilon. $$

Let $M \in \mathbb{N}$ be such that all of the terms $x_1, \cdots, x_N$ are contained as summands in $t_M := y_1 + \cdots + y_M$. It follows that if $m \geq M$, then $t_m - s_n$ is the sum of a finite number of terms $x_k$ with index $k > N$. Hence, for some $q > N$, we have

$$ |t_m - s_n| \leq \sum_{k=N+1}^{q} |x_k| < \varepsilon. $$

Therefore, if $m \geq M$, then we have

$$ |t_m - x| \leq |t_m - s_n| + |s_n - x| < \varepsilon + \varepsilon = 2\varepsilon. $$

Since $\varepsilon > 0$ is arbitrary, we conclude that $\sum y_k$ converges to $x$.

Why couldn't we use normal convergence? I was thinking of the harmonic series as a counterexample, where it does converge (not absolutely) but there is a rearrangement that is divergent, namely: $$ \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \sum_{k=1}^\infty \frac{1}{2k-1} + \sum_{k=1}^\infty \frac{-1}{2k} = \infty - \infty $$

Is this a valid counterexample?

Answer 1

As is pointed in the comments, the Rieman series theorem https://en.wikipedia.org/wiki/Riemann_series_theorem states that a semi convergent series(convergent but not absolutely convergent) can be rearranged in such a way that its sum is any real number(Or +/- $\infty$). As such, this proves that the absolute convergence is necessary for the result to hold. Your counterexample is half valid as you're adding first all the even terms and then all the odd terms which is not a valid rearrangement. The usual counterexample is given by $$1 - 1/2 - 1/4 + 1/3 - 1/6 - 1/8 + ...$$ (taking 1 odd term and 2 even terms), grouping the terms 3 by 3 we get $$(1 - 1/2 - 1/4) + (1/3 - 1/6 - 1/8) + ...$$ which is $$(1/2 - 1/4) + (1/6 - 1/8) + ...$$ which is $$1/2(1 - 1/2 + 1/3 - 1/4 +...)= \log(2)/2.$$ So you changed the sum from log(2) to log(2)/2. In general the way to understand the theorem is to think about the positive terms and the negative terms of the series. We have that the partial sum of the series is the partial sum of the positive terms minus the partial sum of the negative terms. The partial sum of the absolute value is the partial sum of the positive terms plus the partial sum of the negative terms. Then you should convince yourself that the only way for this to be possible is that both the negative terms and the positive terms sum to infinity. Then to construct a rearrangement that goes to a certain real number, you pick up terms from the positive terms of the series until you go over this number (which is always possible since the positive part of the series sums to infinity) and once you are above this number, you pick up terms from the negative part, this processus will enumerate all the indexes. Moreover, since the serie is convergent, the general term goes to 0 and so you must converge to the real number you're aiming for !