Positive and negative portions of conditionally convergent series

Question

I want to prove that for a conditionally convergent series $\sum_{n = 1}^{\infty} a_n$, that its positive subseries (Let $K := \{k \vert a_k > 0\}$, and the positive subseries is $\sum_{j \in K} a_j$) and a (similarly defined) negative subseries diverge.

I know that there is a proof here but I'd want to prove it moreso using definitions. Call $s_n$ the sum of the first n $a_i$, and we know $\exists s$ so for all $n> N$, $\vert s_n - s \vert < \epsilon$. Call $\bar{s}_n$ the sum of the first n $\vert a_i \vert$, and we know $\forall M \exists n$ so $\bar{s_n} >M$.

So since the series $a_n$ is conditionally convergent, there can exist no $N$ so that after that $N$, all terms have the same sign. Thus I want to say something along the lines of, given some $\epsilon$, say 1, there must be a positive subseries of $a_n$ such that $\sum_{i=1}^{n} a_{k_i}$ is above the interval $(s-1,s+1)$, then we can pick do this again with more positive terms so its above the interval $(s-2,s+2)$ and so on. Can I go anywhere with this argument?

Answer 1

I followed you in your first two paragraphs, but I'm not sure I see where you're going in your argument in the third paragraph. I would recommend trying something like the following: First, the positive series and the negative series cannot both converge (since then the difference would converge, which implies that the absolute series converges). So, you just need to handle the case where either the positive series converges and the negative series diverges, or vice versa. Assume without loss of generality that the positive series converges and the negative series diverges. You know that by going out far enough in the series, the sum of the positive terms in the tail can be made arbitrarily small. But then, since the negative series diverges, the tail has to diverge, which implies that the entire series diverges, a contradiction. Thus, both the positive and the negative series must diverge.