Proof
For simplicity, this proof assumes first that an ≠ 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 an positive, with all negative terms replaced by zeroes, and the series includes all an 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 ∑ aσ (n). After the first change of direction, each partial sum of ∑ aσ (n) differs from M by at most the absolute value or of the term that appeared at the latest change of direction. But ∑ an converges, so as n tends to infinity, each of an, and go to 0. Thus, the partial sums of ∑ aσ (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 Ak of integers and a real number Sk are defined. For every k > 0, the induction defines the value σ(k), the set Ak consists of the values σ(j) for j ≤ k and Sk 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 Sk ≤ M, then σ(k+1) is the smallest integer n ≥ 1 such that n is not in Ak and an ≥ 0; if Sk > M, then σ(k+1) is the smallest integer n ≥ 1 such that n is not in Ak and an < 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.
Read more about this topic: Riemann Series Theorem
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