I'm thinking about the following problem:
"Let $\sum_{n=0}^{\infty} a_n$ be a conditionally convergent series;
show that there exists a bijection $f\colon \mathbb{N}\to\mathbb{N}$ such that $\sum_{m=0}^{\infty} a_{f(m)}$ diverges to $+\infty$, or more precisely that $\liminf_{N\to\infty} \sum_{m=N}^{\infty} a_{f(m)}=\limsup_{N\to\infty}\sum_{m=N}^{\infty} a_{f(m)}$".
Intuitively, I think I could make such a rearrangement by choosing an arbitrary $M\in\mathbb{R}^+$, then add positive terms $a_n \geq 0$ (of which there is an infinite number, as I've previously shown) until the sum exceeds $M$, then add one negative term $a_n <0$ then add positive terms until the sum exceeds $2\cdot M$, then add one negative term, and so on.
However, I'm not sure about whether this works and, if it does, how to make it rigorous so I would appreciate any hint/comment/help about how to make this argument correct.
Best regards,
lorenzo.
Best Answer
Indeed the situation is not the same. When you want a finite limit, you alternative overshoot (using the $A_+=\{n:a_n>0\}$ elements) and undershoot (using the $A_-=\{n:a_n\leq 0\}$ elements). In the infinite case you never finish the first overshoot so you will not exhaust the set $A_-$.
Your suggestion, however, is fine. Here is a way to write it out:
Enumerate $A_+$ and $A_-$, calling the sequences $a^+_j$ and $a^-_j$ ($j\geq 1$). Then define inductively: $T_0=0$ and $$ T_M= \inf \left\{ p> T_{M-1} : \sum_{1\leq j\leq p} a^+_j + \sum_{1\leq i<M} a^-_i > M \right\} $$ You first use $T_1$ elements from $A_+$ (to exceed 1), then one element from $A_-$ then $T_2-T_1$ elements from $A_+$ (to exceed 2), next element from $A_-$ etc.. To write down a bijection explicitly, denote:
$$ \phi_+: {\Bbb N} \rightarrow A_+ \ \ \ \mbox{and} \ \ \ \phi_-: {\Bbb N} \rightarrow A_- $$ the monotone bijections with the sets of positive/negative elements. We then define a bijection $ \phi: {\Bbb N} \rightarrow {\Bbb N} $ as follows: For $k\in {\Bbb N}$ of the form $k=T_j+j$ for some $j\geq 1$ we set: $$\phi(k)=\phi_-(j)$$ and for $T_j+j+1\leq k \leq T_{j+1}+j$ for some $j\geq 0$ we set: $$ \phi(k) = \phi_+(k-j)$$ As $T_j$ is monotone in $j$ one verifies that $\phi$ is indeed a bijection and that $$\sum_{k=1}^N a_{\phi(k)} \rightarrow +\infty$$