TITLE: For *any* real sequence, can we make it so that, if we sum parts of it's series in order the terms came in, the resulting sequence is monotone?
QUESTION [5 upvotes]: I was inspired by my own question here, to ask a far more general one:

If $\ (a_n)_{n\in\mathbb{N}}\ $ is a sequence of real numbers, then
does there exist a strictly increasing sequence of natural numbers $\
k_1,\ k_2,\ k_3,\ \ldots\ \quad $ i.e. $\ k_1<k_2<k_3<\ldots\ $ such
that the following sequence,
$$\ \left(\ \displaystyle\sum_{n=1}^{k_1} a_n\ ,\ \sum_{n=k_1+1}^{k_2}
a_n\ ,\ \sum_{n=k_2+1}^{k_3} a_n\ ,\ \ldots\right)\ $$
is a monotone sequence of real numbers.

By monotone, I mean the sequence is either non-increasing or non-decreasing.
I can't think of a counter-example off the top of my head. However, I have no idea how I would go about proving such a result. I think that this would be quite an amazing result if true.
I am thinking that such a counter-example similar to $\ (a_n) = (\ 1,\ 0,\ -1,\ 0,\ 1,\ 0,\ -1,\ 0,\ \ldots)\ $ should exist (i.e. there's no way to split this up in the above requirement in order to make this monotone). Clearly this isn't true for this sequence: $(1+0+-1,\ 0+1+0+-1,\ 0+1+0+-1,\ 0+1+0+-1,\ \ldots) = (0,\ 0,\ 0,\ \ldots)\ $ is monotone. But like I say, I feel like there ought to be an oscillating counter-example related to this one.
If the answer is yes, then I guess we would have to do some sort of proof by contradiction [direct proof seems harder]. "Suppose such a counter-example exists. But such a counter-example cannot exist, because if it did, we get some sort of contradiction."

REPLY [5 votes]: HINT: Consider the sequence
$$
a_n = \begin{cases}
-1 && \text{if } n=1 \\
\frac{1}{2^{n-1}} && \text{if } n>1
\end{cases}
$$