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Consider the equation $$1 + 2 + 3 + 4 + \cdots = - \frac{1}{12},$$ ``proved'' by Ramanujan. One correct way to interpret this is that $\zeta(-1) = - \frac{1}{12},$ where $\zeta(s) = \sum_{n = 1}^{\infty} n^{-s}$ for $\Re(s) > 1$, and $\zeta(s)$ is defined by analytic continuation elsewhere.

I seem to remember being told once that this equation was true in the $p$-adic integers. However, on a moment's reflection this is clearly false; the infinite series does not converge in any $\mathbb{Q}_p$. (I must be misremembering what I was told.)

Is there any argument that an amended version of Ramanujan's statement is true $p$-adically, which does not imitate the usual arguments for $\mathbb{R}$? Is it ``obvious'' that the denominator should only be divisible by the primes 2 and 3?

Thank you!

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Brian Conrad suggested to me that I study the theory of $p$-adic zeta functions. One nice reference for this seems to be "$p$-adic Numbers, $p$-adic Analysis, and Zeta-Functions" by Koblitz. A brief look there doesn't seem to answer my question precisely, but the book has a lot of material that looks interesting and very much related. If no one beats me to it, I will see if I can answer my own question after reading Koblitz's book. –  Frank Thorne Jan 1 '12 at 19:25
Dear Frank, For $p$-adic $\zeta$-functions, I think the best reference is the first half of Katz's 1974 (?) article "$p$-adic $L$-functions via moduli" (or some mild variation of that title) in the Arcata volume. He recalls Euler's summation of $\zeta(-k)$, interprets it $p$-adically, and uses it to prove the existence of the Kubota--Leopold $p$-adic $\zeta$-function, all in five or six pages. It is quite a beautiful exposition. Regards, Matthew –  Emerton Jan 3 '12 at 3:08
Hi Matthew, thank you! I will be sure to look that up as soon as I get back to my library. –  Frank Thorne Jan 4 '12 at 3:58

2 Answers 2

For the primes in the denominator, there is an amusing heuristic based on the fact that $n \equiv n^{-1}\ mod\ p$ holds for all $n\geq 1$ (coprime to $p$) only for $p=2$ and $p=3$. So for these primes the series is like the series $1+1/2+1/3+\cdots$ for $\zeta(1)$, which has a true pole...

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Dear Denis, As you probably know, this is highly related to the discussion of $p$-adic $\zeta$-functions and Kummer congruences in monodromy's answer. Regards, Matthew –  Emerton Jan 3 '12 at 3:06
Sure, but it is easier to remember when one doesn't know much about $p$-adic zeta functions... –  Denis Chaperon de Lauzières Jan 3 '12 at 7:53
+1, thank you, a very interesting observation! After reading the sources recommended to me, I will certainly revisit this fact! –  Frank Thorne Jan 4 '12 at 4:06

First, it seems that this equation was actually first ``proved'' by Euler. In the preface to the book "Elementary theory of $L$-functions and Eisenstein series" by Haruzo Hida, the author gives a beautiful exposition of Euler's manipulations leading to this formula.

The connection to $p$-adic zeta functions seems to be via the Kummer congruences: looking at Euler's formulae, Kummer was apparently lead to the congruences that bear his name, and an appropriate interpretation of the latter by Kubota and Leopoldt half century ago gave rise to the first construction of the $p$-adic analogue of the Riemann zeta function -- the so-called Kubota-Leopoldt $p$-adic zeta function.

Regarding your questions: for the first, I can only say that the zeta values in the Kummer congruences need to be amended to give rise to a continuous function of a $p$-adic variable $s$ (essentially, one needs to remove the corresponding values of the Euler factor $1-p^{-s}$); for the second, I do not see any a priori ``obvious'' reason why $2$ and $3$ should be the only primes in the denominator in Euler's formula.

The first four pages of Pierre Colmez's notes are an excellent reference for the mathematical facts referred to in the last two paragraphs.

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In addition to Hida's preface, at… the longer of my two answers works out the calculations by Euler for the zeta-function at negative integers. –  KConrad Jan 2 '12 at 14:52
Interesting, thank you, I will have a look! –  Frank Thorne Jan 4 '12 at 4:06

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