Riemann rearrangement theorem The Riemann rearrangement theorem states that if $\sum\limits_{n=0} ^{+ \infty} a_n$ is conditionally convergent and $M \in \mathbb{R}$ then there exists a permutation $ \sigma (n) $ such that $\sum\limits_{n=0}^{+ \infty} a_{\sigma(n)} \ =M$.
Could you tell me how to use this to prove a more general statement?
That if we have $\sum\limits_{n=0} ^{+ \infty} c_n$ - conditionally convergent series of complex numbers, then there exists a line $l$ on the plane such that each point of this line can be a limit of the series.
I'd appreciate any help.
 A: A complex series converges if and only if the real and imaginary parts converge, and an identical statements holds when taking absolute values. Then if $\sum_{n=1}^{\infty} c_{n}$ is convergent, so are $\sum_{n=1}^{\infty} a_{n}$ and $\sum_{n=1}^{\infty} b_{n}$ where $c_{n} = a_{n} + ib_{n}$. If $\sum_{n=1}^{\infty} |c_{n}|$ diverges, at least one of $\sum_{n=1}^{\infty} |a_{n}|$ of $\sum_{n=1}^{\infty} |b_{n}|$ diverges, and suppose without loss of generality that only one converges, and that it is the former. Then we can force the real part of our series, $\sum_{n=1}^{\infty} a_{n}$ to converge to whatever we want by the original Riemann Rearrangement Theorem. If $\sum_{n=1}^{\infty} |b_{n}|$ converges, we might be stuck with a fixed sum for the complex part, but we can still hit the entire horizontal line through $i\sum_{n=1}^{\infty} b_{n}$.
If both the complex and real parts are conditionally convergent, the whole situation becomes more complicated...
A: 
This answer follows Ben Webers Generalization of Riemann's Rearrangement Theorem.
The paper has a focus on the rearrangement of conditionally convergent series of complex numbers. It is based upon the concept of a divergent direction within a conditionally convergent series, which is an angle $\theta$ where the points of the series that have an argument close to the angle form a divergent subsequence of $\sum(z_n)$.

Here I cite the propositions and examples in order to give an impression of the central ideas. The rather short proofs are fully stated in the paper.
The paper starts with two central definitions. 

Definition 1: A series of complex numbers $\left(\sum z_n\right)$ is said to converge to a number $a$ if given any $\varepsilon>0$ there exists some $M\in\mathbb{N}$ such that for any $m\geq M$, $\left|\sum_{n=1}^m(z_n)-a\right|<\varepsilon$.
If $\left(\sum|z_n|\right)$ also converges the series is absolutely convergent, conversely if $\left(\sum|z_n|\right)$ diverges, the series is conditionally convergent.
Definition 2: A divergent direction of a conditionally convergent series $\sum(z_n)$ is an angle $\theta$ such that for any $\varepsilon>0$,
  \begin{align*}
  \sum_{\left|\mathop{Arg}(z_n)-\theta(\mathrm{mod}{\;2\pi})\right|<\varepsilon}z_n
  \end{align*}
  is properly divergent. The set of $(z_n)$ that are not part of a divergent direction either are finite, or form an absolutely convergent series.

The goal of the paper is to show the following:
If we denote with $L$ the set of all $c\in\mathbb{C}$ of a complex series $\sum z_n$ for which there is a bijection $\tau:\mathbb{N}\longrightarrow\mathbb{N}$ with $\sum z_{\tau(n)}=c$, then we have always one of the following cases:


*

*$L=\emptyset$ (the series is properly divergent)

*$L$ consists of one point (the series is absolutely convergent)

*$L$ is a line $a+tb$, with $a,b\in\mathbb{C},t\in\mathbb{R}$ (the series is conditionally convergent with two divergent directions)

*$L=\mathbb{C}$ (the series is conditionally convergent with more than two divergent directions).

Lemma 3: If $\sum(z_n)$ is a conditionally convergent series of complex numbers, it has at least two divergent directions.

Example 4: The sequence $(z_n):=\left(\frac{1}{2^n}+\frac{(-1)^{n+1}}{n}i\right)$ has two divergent directions: $\frac{\pi}{2}$ and $\frac{3\pi}{2}$.
Example 5: The sequence $(z_n):=\frac{1}{n}\exp\left(i\pi\frac{n}{2}\right)$ has four divergent directions: $\frac{\pi}{2},\pi,\frac{3\pi}{2}$ and $0$.

Theorem 6: (Rearrangement of series that converge to 0).
  Let $\sum (z_n)$ be a conditionally convergent series of complex numbers that converges to $0$ and let $E$ be a finite set of elements of $(z_n)$. Then there exists a rearrangement $\sigma(n)$ such that $\sum(z_{\sigma(n)}) \setminus E \rightarrow 0$.
Corollary 7:  Removing a finite number of elements from the tail of a conditionally convergent series need not affect its convergence.
Lemma 8: If $(\sum z_n) \rightarrow 0$ and $\mathop{Arg}(c)$ is a divergent direction of $(z_n)$, there
  exists a rearrangement of $(z_n)$ that converges to $c$.
Corollary 9: If  $(\sum z_n) \rightarrow a$ and $\mathop{Arg}(c-a)$ is a divergent direction of $z_n$, then
  there exists a rearrangement of $z_n$ that converges to $c$.

Example 10: Because $\sum (z_n) := \sum_{j=1}^n\left(\frac{1}{2^j}+\frac{(-1)^{j+1}}{j}i\right)$ converges to $1+\ln(2)i$ and $\frac{\pi}{2}$ is a divergent direction, there exists a rearrangement of $(z_n)$ that converges
to $1 + i$ since $\mathop{Arg}(1+\ln(2)i-(1+i))=\mathop{Arg}((\ln(2)-1)i)=\frac{\pi}{2}$.

Lemma 11: If $(\sum z_n) \rightarrow a$ conditionally and has exactly $2$ divergent directions, then those divergent directions are $\theta$ and $-\theta$.
Lemma 12: If $(\sum z_n) \rightarrow a$ and $(\sum z_{\sigma(n)})\rightarrow b$ and $\mathop{Arg}(a-b)$ is not a divergent direction of $\sum (z_n)$, then $\sum (z_n)$ has at least $3$ divergent directions.

The last theorem proves that the set $L=\mathbb{C}$ iff the number of divergent directions is greater than $2$.

Theorem 13: If $(\sum z_n) \rightarrow 0$ and $(z_n)$ has at least three divergent directions then for all $c\in \mathbb{C}$ there exists a rearrangement $z_{\pi(n)}$ such that $\left(\sum z_{\sigma(n)}\right)\rightarrow c$.

