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From Riemann's series theorem, we know that, given a conditionally convergent series, we can permute the elements of the series in order to basically do whatever we want. I have seen a rearrangement of the alternating harmonic series that converges to $\frac{3}{2}\ln{2},\frac{1}{2}\ln{2},\infty$, but I have not seen arrangements that do not converge to anything, that is not even to $\pm\infty$.

My idea was to somehow create an arrangement such that two subsequences of the partial sums that each will tend to a different value, but I was not able to come up with one.

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    $\begingroup$ it is based on a very simple fact : if $a_n \ge 0$ and $b_n \ge 0$ and $\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty b_n = + \infty$ and that $a_n \to 0, b_n \to 0$, then you can approximate any real number with $\sum_{n=A}^N a_n - \sum_{n=B}^M b_n$ by choosing correctly $A,B,N,M$ $\endgroup$ – reuns May 22 '16 at 16:01
  • $\begingroup$ can you use this to prove that a reordering of $\sum_n \frac{(-1)^n}{n}$ can converge to any $x \in \mathbb{R} \cup \{-\infty,+ \infty \}$ ? $\endgroup$ – reuns May 22 '16 at 16:04
  • $\begingroup$ I am looking for a rearrangement that goes nowhere, doesn't blow to $\infty$ and doesn't get close to any real number $\endgroup$ – Joshhh May 22 '16 at 16:06
  • $\begingroup$ if you can approximate any number, then you can approximate $1$, then $-1$, then $1$, then $-1$... and hence the series doesn't converge : so you didn't find any example because it obvious once you proved the general case $\endgroup$ – reuns May 22 '16 at 16:08
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If $\sum a_n$ is a conditionally convergent series of real numbers and $\alpha\le\beta$ then there is a rearrangement so that $\liminf s_n=\alpha$ and $\limsup s_n=\beta$.

The proof is the same as the proof that there is a rearrangement converging to $\alpha$. Start with just enough positive terms to give a partial sum larger than $\beta$, then add just enough negative terms to give a partial sum less than $\alpha$, then add more positive terms til you get a partial sum larger than $\beta$ again, etc. See the proof of the rearrangement theorem that you mention for details.

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Fix two real number $\alpha,\beta $ and assuming $\alpha>\beta $. Since $\sum_{n\geq1}\frac{1}{2n} $ diverges, we can find some $N_{1} $ such that $$\alpha<\sum_{n\leq N_{1}}\frac{1}{2n}:=S $$ and now since $\sum_{n\geq1}\frac{1}{2n+1} $ diverges we can find some $N_{2} $ such that $$\beta>S-\sum_{n \leq N_{2}}\frac{1}{2n+1} $$ and so on. So $\alpha,\beta $ are accumulation point of the partial sums.

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In fact, if $\sum a_n$ is conditionally convergent, then there is a rearrangement of $\sum a_n$ such that the partial sums of the rearranged series form a dense subset of $\mathbb R.$

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