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If we naïvely apply the formula $$\sum_0^\infty a^i = {1\over 1-a}$$ when $a=2$, we get the silly-seeming claim that $1+2+4+\ldots = -1$. But in the 2-adic integers, this formula is correct.

Surely this is not a coincidence? What is the connection here? Do the $p$-adics provide generally useful methods for summing divergent series? I know very little about either divergent series or $p$-adic analysis; what is a good reference for either topic?

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What is the coincidence? The argument that justifies the convergence of geometric series applies in any complete normed field when $|a| < 1$, and in the $2$-adics this is true when $a = 2$. – Qiaochu Yuan May 6 '12 at 23:02
What exactly is your question? In any case, note that "divergent series" is closely related to the used norm. In the usual setting, $|2^{100}| = 2^{100}$ is large, but in a $2$-adic setting, $|2^{100}|_2 = 2^{-100}$ is small. So in a $2$-adic setting, this is really a convergent series. – TMM May 6 '12 at 23:05
Relevant – Alex Becker May 6 '12 at 23:06
up vote 19 down vote

A counterexample may be interesting. Consider the sequence

$$ a_n = \frac{n!}{n! + 1} $$

In the real numbers, we obviously have

$$ \lim_{n \to +\infty} a_n = 1, $$

but in every $p$-adic field, we have

$$ \lim_{n \to +\infty} a_n = 0, $$

so you should be wary of the idea of trying to sum a series of real numbers by transplanting it to the $p$-adics.

One thing you can consider is $\mathbb{Q}((x))$, the field of rational (formal) Laurent series. In this field, you have an identity

$$ \sum_{n=0}^{+\infty} x^n = \frac{1}{1-x}. $$

There is no issue of convergence or anything here; you just check that multiplying the left hand side by $1-x$ gives you $1$.

There is a subfield of $\mathbb{Q}((x))$ that consists of only those Laurent series for which replacing $x$ by $2$ yields a convergent $2$-adic sum. Evaluation at $2$ then becomes a field homomorphism to the $2$-adic numbers. Since $\sum_{n=0}^{+\infty} x^n$ is in that subfield, its image in $\mathbb{Q}_2$ must be the same as the image of $1/(1-x)$: i.e. $-1$.

Wikipedia has a page on divergent series which talks about "summation methods". You may find this another useful starting point.

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Yes, roughly speaking, the convergence of a sequence or series of rationals in the real sense is completely independent of its convergence $p$-adically. – Lubin May 22 '12 at 19:22
I wish I could upvote this a second time. – MJD Aug 11 '12 at 12:45
what does mean $\lim_{n \to \infty} \frac{n!+1}{n} = 0$ in the p-adic field ? – user1952009 Jan 23 at 23:59
@MJD I just did it for you. ;-) – Stéphane Gourichon Jan 25 at 9:54

This is not so much an answer as a related reference. I wrote a short expository note "Divergence is not the fault of the series," Pi Mu Epsilon Journal, 8, no. 9, 588-589, that discusses this idea and its relation to 2's complement arithmetic for computers.

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As for references, Hardy wrote a book, Divergent Series, which seems to be available for download on the web. There are several books about the $p$-adics; Koblitz, $p$-adic Numbers, $p$-adic Analysis, and Zeta-Functions; Bachman, Introduction to $p$-adic Numbers and Valuation Theory; Gouvea, $p$-adic Numbers. There's also an article I wrote with Alf van der Poorten, Some problems concerning recurrence sequences, Amer. Math. Monthly 102 (1995) 698-705, MR 97a:11029.

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up vote 2 down vote accepted

Qiaochu Yuan points out that it is not at all a coincidence: “The argument that justifies the convergence of geometric series applies in any complete normed field when $|a|<1$, and in the 2-adics this is true when $a=2$.” In the 2-adics, the sequence does indeed converge, and the only thing it can converge to is $1\over1-r$.

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