Divergent series and $p$-adics 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?
 A: 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.
A: 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 $1\over1-r$.
A: 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. 
A: 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.
