Real analysis: Characteristic property for unconditional divergence A convergent series $\sum_{k=1}^\infty a_k$ is called unconditional convergent, when it's value is invariant under any permutation $\sigma:\mathbb N\to\mathbb N$ of it's summands, i.e. $\sum_{k=1}^\infty a_{\sigma(k)}$ converges and $\sum_{k=1}^\infty a_k = \sum_{k=1}^\infty a_{\sigma(k)}$. For real valued series we know:
$$\sum_{k=1}^\infty a_k\text{ is unconditional convergent} \iff \sum_{k=1}^\infty a_k \text{ converges absolutely}$$
First Question: Is there also something like unconditional divergence for series studied in mathematics? It may be defined as

A series is $\sum_{k=1}^\infty a_k$ diverges unconditionally, iff $\sum_{k=1}^\infty a_k$ diverges and for each permutation $\sigma:\mathbb N\to\mathbb N$ also $\sum_{k=1}^\infty a_{\sigma(k)}$ diverges.

For example $\sum_{k=1}^\infty 1$ or $\sum_{k=1}^\infty k$ diverge unconditionally.
Second Question: If it is already studied in mathematics: What is the characteristic property of unconditional divergence for real valued series? It cannot be $\sum_{k=1}^\infty |a_k|=\infty$, because a divergent rearrangement of $\sum_{k=1}^\infty (-1)^k \tfrac 1k$ would be a counterexample...
 A: First of all, if $\lim_{n\to\infty}a_n\neq 0$, the series $\sum a_n$ is unconditionally divergent.

Since $\lim_{n\to\infty}a_n\neq 0$, there is some value $c\in\Bbb R_{>0}$ such that $|a_n|>c$ for arbitrarily large $n$. Suppose $\sum a_n$ converges to some value $x$, then there is some $N\in\Bbb N$ such that all partial sums $S_m=\sum_{n<m} a_n$ with $m>N$ are within a radius less than $\frac 12 c$ of $x$. Take $k\in\Bbb N$ such that $k>N$ and such that $|a_{k+1}|>c$, then $|S_{k}-S_{k+1}|=|a_{k+1}|>c$, contradicting that $|S_m-x|<\frac12c$ for all $m>N$.

Let $a_n$ be a sequence of real numbers, and let $a^+_n$ be the sequence in which all negative terms are replaced by $0$, and similarly let $a^-_n$ be the sequence of which all positive terms are replaced by $0$. Clearly $\sum a^+_n$ is the sum of the positive terms of $a_n$, and $\sum a^-_n$ the sum of its negative terms.
If one of $\sum a^+_n$ and $\sum a^-_n$ is finite and the other tends to infinity, then $\sum a_n$ diverges unconditionally.

Without loss of generality, let $\sum a^+_n$ diverge to $\infty$ and $\sum a^-_n=c\in\Bbb R$. Note that $\sum a^-_n$ is absolutely convergent. Suppose $\pi:\Bbb N\to\Bbb N$ is a permutation.
Since $\sum a^+_n=\infty$, for any $x\in \Bbb R$ we can find some $N\in\Bbb N$ such that $\sum_{n<N}a^+_n>x$. By taking $M\in\Bbb N$ large enough such that $\pi^{-1}(n)<M$ for all $n<N$, we see that $\sum_{n<M}a^+_{\pi(n)}>x$ as well, as all terms are positive.
Since $\sum a^-_n$ is absolutely convergent, $\sum a^-_{\pi(n)}=c$ as well. But then we can clearly find arbitrarily large partial sums, since if we choose $M$ as before such that $\sum_{n<M} a^+_{\pi(n)}>x$, then we have $\sum_{n<M} a_{\pi(n)}>x-c$.

However, if $\lim_{n\to\infty} a_n=0$ and both $\sum a^+_n$ and $\sum a^-_n$ tend to infinity, then there exists a permutation $\pi:\Bbb N\to\Bbb N$ such that $\sum a_{\pi(n)}$ converges.

In fact the series can be permuted to converge to any value in $\Bbb R$, to diverge by oscillation or to diverge by tending to either of the infinities. The proof is as in Riemann's rearrangement theorem.


Therefore we can conclude that a series is unconditionally divergent if and only if its terms do not vanish or if exactly one of the sums of its positive terms or of its negative terms tends to infinity.
In case the terms vanish and both the sums of its positive and of its negative terms tend to infinity, then the series is either conditionally convergent or conditionally divergent, and in case the terms vanish and both sums converge, then the series is unconditionally convergent.
A: An unconditionally divergent series in the sense you described is simply a series that tends to infinity. It's easy to see that a series whose partial sums tend to infinity is unconditionally divergent; conversely, a series which diverges but doesn't tend to infinity must oscillate, and this can be used to show that it can be made convergent by an appropriate permutation of its terms. 
See On Divergent Series by A. S. Chessin, Bull. Amer. Math. Soc. Volume 2, Number 3 (1895), 72-75.
