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Grandi's series appears to have different values depending on how you bracket the terms, e.g. $(1 - 1) + (1 - 1) + (1 - 1) + ... = 0$, while $1 + (-1 + 1) + (-1 + 1) + ... = 1$. This isn't a problem today because we define the sum of a series as the limit of its partial sums (which is this case just doesn't exist). And the observation is that you can't perform rearrangements or even regroupings of a series' terms (unless it is absolutely convergent).

I often see p-adic series being manipulated in similar ways, e.g. equation 2.1 here, where: $\overline{n_0n_1 \dots n_{k-1}} = 1(n_0n_1 \dots n_{k-1}) + p^k (n_0n_1 \dots n_{k-1}) + p^{2k}(n_0n_1 \dots n_{k-1}) + \dots = (n_0n_1 \dots n_{k-1}) ( 1 + p^k + p^{2k} + \dots)$.

I've been trying to convince myself this is safe/meaningful. Looking at partial sums (or really, the values in $\varprojlim \mathbb Z/ p^n \mathbb Z$) I can convince myself the left side equals the right side. But I'm not sure how to formally interpret the middle as a p-adic number (other than just rearranging it to be either the left or right side). I know of p-adic numbers as either formal infinite series of the form $\sum_v a_v p^v$ or elements in $\varprojlim \mathbb Z/ p^n \mathbb Z$. (Or as cauchy sequences).

Should I just think of the middle step as being for intuition?

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    $\begingroup$ You could consider the completion $\mathbb{Q}_p$ of the field of rationals equipped with the $p$-adic valuation. $\endgroup$ Mar 9, 2020 at 5:00
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    $\begingroup$ In $\Bbb{R}$ rearranging is allowed only for absolutely convergent series. In $\Bbb{Q}_p$ there is no such problem: iff $|a_n|_p\to 0$ then $\sum_n a_n$ converges and it doesn't depend on the order of summation. $\Bbb{Z}_p$ is the completion of $\Bbb{Z}$ for the $|ap^k|_p=p^{-k},p\nmid a$ absolute value. A sequence of integers converges iff it converges modulo every $p^k$. $\endgroup$
    – reuns
    Mar 9, 2020 at 5:31
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    $\begingroup$ The way I learned the fact stated by @reuns: In an ultrametric field, $a_n \to 0 \Leftrightarrow \sum a_n$ is absolutely convergent $\Leftrightarrow \sum a_n$ is convergent. Try to prove that yourself via i $\Rightarrow$ ii $\Rightarrow$ iii $\Rightarrow$ i. -- The middle step in your calculation makes total sense as a convergent infinite sum. $\endgroup$ Mar 9, 2020 at 17:20
  • $\begingroup$ Ah I see, so thinking of it as a sequence is the way to go. Many thanks! $\endgroup$
    – vacant
    Mar 9, 2020 at 19:17
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    $\begingroup$ @Merosity: By "absolutely convergent" I mean that for every bijection $\mathbb N \rightarrow \mathbb N$, the sequence of partial sums of $\sum a_{s(n)}$ converges to the same value, vulgo "we can reorder the sum any way we like"; not the criterion of convergence of the sum of the absolute values in $\mathbb R$ that is usually called that in real analysis (which is useless in ultrametric analysis as your example shows). Maybe there is a better word for what I mean, sorry for being unclear there. $\endgroup$ Mar 10, 2020 at 1:42

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