Riemann Series Theorem - Proof

Proof

For simplicity, this proof assumes first that a_n ≠ 0 for every n. The general case requires a simple modification, given below. Recall that a conditionally convergent series of real terms has both infinitely many negative terms and infinitely many positive terms. First, define two quantities, and by:

That is, the series includes all a_n positive, with all negative terms replaced by zeroes, and the series includes all a_n negative, with all positive terms replaced by zeroes. Since is conditionally convergent, both the positive and the negative series diverge. Let M be a positive real number. Take, in order, just enough positive terms so that their sum exceeds M. Suppose we require p terms – then the following statement is true:

This is possible for any M > 0 because the partial sums of tend to . Discarding the zero terms one may write

Now we add just enough negative terms, say q of them, so that the resulting sum is less than M. This is always possible because the partial sums of tend to . Now we have:

Again, one may write

with

Note that σ is injective, and that 1 belongs to the range of σ, either as image of 1 (if a₁ > 0), or as image of m ₁ + 1 (if a₁ < 0). Now repeat the process of adding just enough positive terms to exceed M, starting with n = p + 1, and then adding just enough negative terms to be less than M, starting with n = q + 1. Extend σ in an injective manner, in order to cover all terms selected so far, and observe that a ₂ must have been selected now or before, thus 2 belongs to the range of this extension. The process will have infinitely many such "changes of direction". One eventually obtains a rearrangement ∑ a_σ (n). After the first change of direction, each partial sum of ∑ a_σ (n) differs from M by at most the absolute value or of the term that appeared at the latest change of direction. But ∑ a_n converges, so as n tends to infinity, each of a_n, and go to 0. Thus, the partial sums of ∑ a_σ (n) tend to M, so the following is true:

The same method can be used to show convergence to M negative or zero.

One can now give a formal inductive definition of the rearrangement σ, that works in general. For every integer k ≥ 0, a finite set A_k of integers and a real number S_k are defined. For every k > 0, the induction defines the value σ(k), the set A_k consists of the values σ(j) for j ≤ k and S_k is the partial sum of the rearranged series. The definition is as follows:

• For k = 0, the induction starts with A₀ empty and S₀ = 0.

• For every k ≥ 0, there are two cases: if S_k ≤ M, then σ(k+1) is the smallest integer n ≥ 1 such that n is not in A_k and a_n ≥ 0; if S_k > M, then σ(k+1) is the smallest integer n ≥ 1 such that n is not in A_k and a_n < 0. In both cases one sets

It can be proved, using the reasonings above, that σ is a permutation of the integers and that the permuted series converges to the given real number M.