On the possible values of $\sum\varepsilon_na_n$, where $\varepsilon_n=\pm1$ (i.e., changing signs of the original series)

I have used the following result in an answer on this site.

Suppose that $a_n>0$ are positive real numbers such that $\sum\limits_{n=1}^\infty a_n=+\infty$ and $\lim\limits_{n\to\infty} a_n=0$. Ten for any choice of $A,B\in\mathbb R\cup\{\pm\infty\}$ such that $A\le B$ there exists a sequence $\varepsilon_n$ such that each $\varepsilon_n\in\{\pm1\}$, $$\liminf\limits_{n\to\infty} \sum_{k=1}^n \varepsilon_k a_k=A \qquad\text{ and }\qquad \limsup\limits_{n\to\infty} \sum_{k=1}^n \varepsilon_k a_k=B.$$

In the other words, this says how we can change the behavior of the series by changing the signs of the terms of the series.

Sketch of the proof. We choose two sequences such that $A_n\to A$, $B_n\to B$. We construct the sequence $\varepsilon_n$ in blocks.

In the first block we choose $\varepsilon_k=1$ and we stop at the first $n_0$ when $\sum\limits_{k=1}^{n_1} \varepsilon_ka_k > A_1$.

The second block will have negative signs, and we stop when $\sum\limits_{k=1}^{n_2} \varepsilon_k a_k < B_1$.

By induction we can continue in choosing blocks with $+1$'s and $-1$'s like this. (I.e. we always have $\sum\limits_{k=1}^{n_{2l-1}} \varepsilon_ka_k > A_l$ and $\sum\limits_{k=1}^{n_{2l}} \varepsilon_ka_k < B_l$; moreover, the numbers $n_l$ are always chosen in the $l$-th step of the construction as the minimal possible number with these properties.)

The divergence of the original series is used to get that each block ends somewhere. The fact that $a_n\to0$ is used to get that the values of $\liminf$ and $\limsup$ are indeed $A$ and $B$. $\hspace{2cm}\square$

I think that this result is occasionally useful.

What I am asking for is some reference for this result. (I searched a little and did not find one. I did not find a duplicate question on this site, either.) Or any link to a place, where this result is explained in more detail. (Of course, if you wish to do so, you can post a more detailed proof, different proof or some comments related to this result as an answer.)

Solution 1:

I think Hugo Steinhaus, Jean-Pierre Kahane, and some others have worked on things related to this.

Try googling "conditionally convergent" along with things like "random series" and/or "Kahane" and/or "Steinhaus".

See also Miller/Schnitzer's 1996 paper Measure and category—some non-analogues.