Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
9  
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
2  
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
2  
Relevant –  Alex Becker May 6 '12 at 23:06

4 Answers 4

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.

share|improve this answer
1  
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
1  
I wish I could upvote this a second time. –  MJD Aug 11 '12 at 12:45

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.

share|improve this answer

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.

share|improve this answer
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 $1\over1-r$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.