Rearrangement and Rearrangement Theorem

Rearrangement

Given $\sum_{n = 1}^{\infty} a_{n}$ and a bijection $f : N \to N$ . Define $b_{f (k)} = a_{k}$ . Then $\sum_{n = 1}^{\infty} b_{n}$ is a rearrangement of $\sum_{n = 1}^{\infty} a_{n}$ .

Rearrangement Theorem

If $\sum a_{n}$ is conditionally convergent, and $α \in R$ . Then $\exists$ a rearrangement $\sum b_{k}$ such that $\sum_{k = 1}^{\infty} b_{k} = α$ .

Proof
Think of a proof outline for the AST. If you only add the positive terms, it must diverge, and similarly for the negative terms. Similarly, in a conditionally convergent series you can use the positive terms, add one negative term, then add positive terms until you get too far, then add one negative term, and rinse and repeat:

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Theorem

If $\sum a_{n}$ is absolutely convergent, then any rearrangement converges to the same value.

Proof
Let $f : N \to N$ where $b_{f (k)} = a_{k}$ so that $\sum b_{k}$ is a rearrangment. Then let:

s_{n} = a_{1} + \dots + a_{n}

t_{m} = b_{1} + \dots + b_{m}

Since $\sum a_{n}$ is absolutely convergent, then if we let $ε > 0$ to show convergence then:

\sum a_{n} = A

By our supposition. We can choose $N \in N$ such that for any $n \geq N$ that:

| s_{n} - A | < \frac{ε}{2}

Similarly by the Cauchy Criterion, then $\forall n > m \geq N$ then:

| a_{m + 1} | + \dots + | a_{n} | < \frac{ε}{2}

Let $M = max (f (1), f (2), \dots, f (N))$ . If we choose $m \geq M$ then $m \geq f (i)$ so then:

| t_{m} - A | \leq \underset{\sum a_{k} : k > N}{\underset{⏟}{| t_{m} - s_{N} |}} + \underset{< \frac{ε}{2}}{\underset{⏟}{| s_{N} - A |}}

But since the Cauchy Criterion makes the first term $< \frac{ε}{2}$ , finishing the proof.
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