Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is a theorem of Riemann to that effect. How to prove it?

Note: This was asked by Kenny in the beta for "calculus".

share|cite|improve this question
up vote 1 down vote accepted

More generally, if $\sum\limits_{i=1}^{\infty} c_n$ is a conditionally convergent series of complex numbers, then the collection of sums of all convergent rearrangements is an affine subspace of the plane $\mathbb{C}$.

Here's a sketch of why this is true.

Suppose $A$ and $B$ are two rearrangements that converge to the distinct complex numbers $a,b$, respectively. Focus on a partial sum of $A$ that's pretty close to $a$. A sufficiently long partial sum of $B$ will include every term of the former partial sum of $A$, but it will also tend to $b$ as you take more and more terms. It's "clear" then that it's possible to obtain any combination $ta + (1 - t)b$ for $t \in \mathbb{R}$.

If in addition there is a point $c$ which is the sum of some rearrangement but is not on the line through $a$ and $b$, then the previous result can be reapplied, replacing the numbers $a$ and $b$ with the numbers $c$ and $d$, where $d$ is any point on the line through $a$ and $b$. It follows that in this case the entire plane $\mathbb{C}$ is obtained by rearranging the original series.

share|cite|improve this answer

Let the number be r. The idea is that you first add all the positive elements of the sequence in order, until you get over r, then you add all the negative elements until you get under r, and then all the positive until you get over r, and so forth.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.