convergence divergence - Infinite series of positive and negative terms both divergent

Let $x$ be the infinite series

$$\sum_{k=0}^\infty f(k)$$

Let $p$ be the sum of all positive terms and $n$ the sum of all negative terms from $x$. My question is: if $p$ and $n$ are both divergent, does $x$ have no absolute sum? Can it be rearranged to form any arbitrary resulting value? I believe it is true because if at least one of $p$ and $n$ was convergent, then $x$ would have an absolutely determinable sum.

Answer

In a word, yes. In more detail, let $(a_{k})_{k=0}^{\infty}$ be a real sequence, and put

\begin{align*}
a_{k}^{+} &= \phantom{-}\max(a_{k}, 0) = \tfrac{1}{2}\bigl(|a_{k}| + a_{k}\bigr), \\
a_{k}^{-} &= -\min(a_{k}, 0) = \tfrac{1}{2}\bigl(|a_{k}| - a_{k}\bigr),
\end{align*}
so that
$$
|a_{k}| = a_{k}^{+} + a_{k}^{-},\qquad
a_{k} = a_{k}^{+} - a_{k}^{-}.
$$
The following are equivalent:

1. $\sum\limits_{k=0}^{\infty} |a_{k}|$ converges (i.e., $(a_{k})_{k=0}^{\infty}$ is absolutely summable);




2. Each of $\sum\limits_{k=0}^{\infty} a_{k}^{\pm}$ converges (absolutely).

Further,

3. If $\sum\limits_{k=0}^{\infty} a_{k}$ converges conditionally (i.e., $(a_{k})_{k=0}^{\infty}$ is summable, but not absolutely), then each of $\sum\limits_{k=0}^{\infty} a_{k}^{\pm}$ diverges.

Proofs:

1. implies 2.: Comparison, plus $a_{k}^{\pm} = |a_{k}^{\pm}| \leq |a_{k}|$.

2. implies 1.: $|a_{k}| = a_{k}^{+} + a_{k}^{-}$.

3. If both $(a_{k})$ and $(a_{k}^{-})$ are summable, then so is $(a_{k}^{+})$, since $a_{k}^{+} = a_{k} + a_{k}^{-}$, so $(a_{k})$ is absolutely summable since $|a_{k}| = a_{k}^{+} + a_{k}^{-}$. Contrapositively, if $(a_{k})$ is summable but not absolutely summable, then $(a_{k}^{-})$ is not summable.

An entirely similar argument shows that if $(a_{k})$ is conditionally summable, then $(a_{k}^{+})$ is not summable.

As you say, if $(a_{k})$ is conditionally summable, the series can be rearranged to converge to an arbitrary sum (or so that the partial sums diverge "in a more-or-less arbitrary way").