# Riemann's theorem for complex series

1. Mar 24, 2010

### zpconn

Suppose the real numbers {a_k} are the terms of a conditionally convergent series. Let C({a_k}) denote the set of all sums of rearrangements of this series. A famous theorem of Riemann shows that, in this particular circumstance, C({a_k}) = R, the set of real numbers.

Various generalizations of this result exist for Banach spaces, but I'm interested in the special case of complex series. In particular, if the complex numbers {a_k} are the terms of a conditionally convergent series, then C({a_k}) is an affine subspace of the complex plane.

I was told this informally by a professor, but the result is completely fascinating to me, so I've been trying to find a proof, either online or by working one out myself. Well, so far I have failed in both endeavors!

I was wondering if anyone had any insight into this problem, either a reference or an idea on how to proceed in proving it.

One thing to note is that given an affine subspace of the plane, one can construct a series with terms {a_k} such that C({a_k}) is that particular subspace. For instance, if {alpha,beta} is a basis of the complex plane and {a_k} and {b_k} are *real* numbers that are the terms of two conditionally convergent series, then C({alpha * a_k + beta * a_k}) is the plane by Riemann's theorem as stated above for series with real terms.

In short, it's evident that any affine subspace can be achieved in the form C({a_k})--the significant part is that C({a_k}) must always have this form! (Of course, this is provided that the associated series is conditionally convergent.)

2. Mar 25, 2010

### Eynstone

A convergent rearrangement of a_k determines that of b_k.The latter may not converge. Although your guess is right, we can't work with individual components.
Here's a lemma which will prove the affinity of the subspace.
Lemma : Let {c_n} be a conditionally convergent series ,with two rearrangements A & B converging to distinct complex numbers a,b. Then the series can be made to converge to ka+(1-k)b for any real number k.
Here's why it works. Select a partial sum of A, which is close to a. The partial sums of B which include this partial sum eventually tend to b . Thus, we can 'move' along the direction a-->b ( back & forth by repeted selection) & generate the line.
Suppose that complex number c not on this line is in C{c_n}. We can apply the lemma once more to infer that the range is ka+lb+mc for any real numbers k,l,m with
k+l+m =1. This subspace is clearly the complex plane. (Essentially, if c c joined to every point on line a--b, the plane is spanned).
The generalisation to real vector spaces follows similarly.