conditionally convergent but not absolutely convergent series

Question

I'm stuck on the following exercise:

Let $\sum_{n=0}^{\infty} a_n$ be a series of real numbers which is conditionally convergent, but not absolutely convergent.

Define the sets $$A_+:=\{n\in\mathbb{N}:a_n \geq 0\}$$ and $$A_-:=\{n\in\mathbb{N}:a_n<0\},$$ thus $$A_+ \bigcup A_-=\mathbb{N}\ \text{and}\ A_+ \bigcap A_-=\emptyset.$$ Then both of the series $\sum_{n\in A_+}a_n$ and $\sum_{n\in A_-} a_n$ are not absolutely convergent.

$\sum_{n\in A_+} a_n$ and $\sum_{n\in A_-} a_n$ can't be both absolutely convergent at the same time is straightforward since it follows that $\sum_{n\in A_+\bigcup A_-}a_n =\sum_{n\in\mathbb{N}} a_n$ is absolutely convergent, a contradiction.

What I haven't been able to do to is exclude the possibility that one of the two converges and the other diverges, i.e. that the remaining two cases:
(1) $\sum_{n\in A_+} a_n$ absolutely convergent, $\sum_{n\in A_-} a_n$ not absolutely convergent;

(2) $\sum_{n\in A_+} a_n$ not absolutely convergent, $\sum_{n\in A_-} a_n$ absolutely convergent;

it leads to a contradiction.

So, I would appreciate any hints about how to carry out this part of the proof.

Best regards,

Lorenzo.

Answer 1

Let $b_n = (|a_n| + a_n)/2$ and $c_n = (|a_n|-a_n)/2.$

Then the partial sums satisfy

$$\sum_{n=1}^m a_n = \sum_{n=1}^m b_n - \sum_{n=1}^m c_n, \\ \sum_{n=1}^m |a_n| = \sum_{n=1}^m b_n + \sum_{n=1}^m c_n.$$

If $\sum a_n$ converges and $\sum |a_n|$ diverges, then both $\sum b_n$ and $\sum c_n$ diverge, since

$$2\sum_{n=1}^m b_n = \sum_{n=1}^m |a_n| + \sum_{n=1}^m a_n, \\ 2\sum_{n=1}^m c_n = \sum_{n=1}^m |a_n| - \sum_{n=1}^m a_n,$$

and the sum or difference of a divergent and convergent series is divergent.

Furthermore, we have divergence to $+\infty$ in each case, as the partial sums of $|a_n|$ form a non-negative, non-decreasing sequence.

Note that

$$\{b_n: n \in \mathbb{N}, b_n \neq 0\} = \{a_n: n \in A^+, a_n \neq 0\}, \\ \{c_n: n \in \mathbb{N}, c_n \neq 0\} = \{-a_n: n \in A^-\}, $$ and it easily shown that

$$ \sum_{n\in A_+} a_n=\sum_{n=1}^\infty b_n = +\infty\\ \sum_{n\in A_-} a_n = -\sum_{n=1}^\infty c_n = - \infty $$

Answer 2

Notice that

$$\sum_{n=0}^\infty a_n=\sum_{n\in A_+}\vert a_n\vert - \sum_{n\in A_-}\vert a_n\vert$$

So if either of them converges absolutely, so does the other since $\sum_{n=0}^\infty a_n<\infty$