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.
Best Answer
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...