Is the understanding of the proof of rearrangement theorem correct

absolute-convergenceproof-explanationreal-analysissequences-and-series

I'm reading the proof of rearrangement theorem. Could you please verify if my understanding of the last part of the proof is correct or not?

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  1. The authors said

The inequality $(8.3)$ implies the absolute convergence of $\sum x_{\sigma(k)}$.

I think the inequality $(8.3)$ implies that $\sum_{k=0}^{m}|x_{\sigma(k)}| \le \sum_{k=0}^{N}|x_{k}| + \varepsilon$ for all $m \ge M$. As such, the sequence $(\sum_{k=0}^{m}|x_{\sigma(k)}|)_{m \in \mathbb N}$ is bounded and thus it converges. Since $(\sum_{k=0}^{m}|x_{\sigma(k)}| )_{m \in \mathbb N}$ converges, $\sum x_{\sigma(k)} := (\sum_{k=0}^{m} x_{\sigma(k)})_{m \in \mathbb N}$ converges absolutely.

  1. The authors said

We see that $\left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{N} x_{k}\right| \leq \varepsilon$, so the values of the two series agree.

I think because $\left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{N} x_{k}\right| \leq \varepsilon$ holds, $\left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{n} x_{k}\right| \leq \varepsilon$ holds for all $n \ge N$. As such, we take the limit $n \to \infty$ and get $$\left|\sum_{k=0}^{\infty} x_{\sigma(k)}- \lim_{n \to \infty}\sum_{k=0}^{n} x_{k}\right| \leq \varepsilon \quad \text{or equivalently} \quad \left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{\infty} x_{k}\right| \leq \varepsilon$$

We again take the limit $\varepsilon \to 0$ and get $$\left|\sum_{k=0}^{\infty} x_{\sigma(k)}-\sum_{k=0}^{\infty} x_{k}\right| \le \lim_{\varepsilon \to 0}\varepsilon = 0 \quad \text{or equivalently} \quad \sum_{k=0}^{\infty} x_{\sigma(k)} = \sum_{k=0}^{\infty} x_{k}$$

As such, the values of the two series agree.

Best Answer

  1. Correct.

  2. Let $S:=\sum_{k=0}^\infty x_{\sigma(k)}$. Then $\big| S-\sum_{k=0}^n x_k\big|\le\varepsilon$ will indeed follow for $n>N$, but your reasoning is not enough.
    Use that, just as for $N$, there's also an $M'(\ge M)$ for $n$, such that all indices $0,1,\dots,n$ are present within $\sigma(0),\dots,\sigma(M')$.

    Once it's proved, we can simply finish by the definition of limit, as for every $\varepsilon>0$ we have an $N$ such that $n\ge N$ implies $|S-\sum_{k=0}^n x_k|\le\varepsilon$.

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