## Summation methods for divergent series

There are many methods for assigning a value to a series that diverges, e.g. zeta function regularization, Abel summation, Cesaro summation, etc. From all of the examples I've found, two methods either give the same result or one of them doesn't work. For example, both zeta function regularization, Ramanujun summation, and a method of Euler assign -1/12 to 1 + 2 + 3 + 4 + ... while Abel summation can't assign a value. My question is if there is an example of a series that different summation methods assign different values to or is it the case that any two summation methods must agree on divergent series (that they can assign a value to). Here I am assuming that both summation methods assign the correct value to convergent series and are linear. I am guessing we need stronger conditions since it seems that the space of convergent series is not dense, in some sense, in the space of all series.

EDIT: I was able to pick up a copy of Hardy's "Divergent Series." It's a really neat book but I have yet to be able to find in it an example of a divergent series that gets assigned two different values by two different linear and consistent summation methods. He does show how a method being linear forces specific series to have a unique value. Surely the issue of whether two general summation methods (with reasonable conditions that they satisfy) can disagree on a certain series must come up somewhere in the literature.

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Related: mathoverflow.net/questions/3204/… – Qiaochu Yuan Mar 24 2010 at 19:26
Also related: three posts at The Everything Seminar: cornellmath.wordpress.com/2007/07/28/…, cornellmath.wordpress.com/2007/07/30/…, cornellmath.wordpress.com/2007/08/02/… – Tom Leinster Mar 25 2010 at 3:47

If a series has a well-defined Cesaro sum, then it has a well-defined Abel sum and they are equal. I think I first learned this from Hardy's Divergent Series; the proof is short enough to give here.

Let $a_i$ be the series in question, let $s_m = \sum_{i=0}^m a_i$ and $c_n = \sum_{m=0}^n s_m$. The claim that the Cesaro sum is well defined is that $$c_n = (n+1)(L + o(1)).$$

Let $A(x) = \sum a_i x^i$. Then $\sum c_i x^i = A(x)/(1-x)^2$ so $$A(x) = (1-x)^2 \sum_{n=0}^{\infty} (L (n+1) + o(n+1)) x^n$$ where the $o$ is as $n \to \infty$, independent of $x$. But $$\sum_{n=0}^{\infty} (n+1) x^n = 1/(1-x)^2$$ so $$A(x) = L (1-x)^2/(1-x)^2 + o \left( (1-x)^2/(1-x)^2\right) = L+ o(1) \quad \mbox{as} \ x \to 1^{-}.$$ So the Abel sum of $a_i$ is also $L$.

Deleted an argument that Cesaro summability implies zeta summability; not sure I can sum by parts where needed.

I want to say that I feel guilty writing up special cases like this. I have a vague impression that there is a very general philosophy here, something like Wiener's generalized Tauberian theorem. (But presumably easier, since we are generalizing Abel's theorem, not Tauber's.) I'm hoping that someone will come by and write up an exposition of it.

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The series 1-1+1-1+... has a long history of people assigning it different values using different methods.

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I think it is nowadays accepted that the "correct" value, such as it is, is 1/2. Is there any modern contention of this value? – Qiaochu Yuan Mar 25 2010 at 0:08
I agree with Qiaochu -- certainly some people have claimed that its value is 0, and others (sometimes even the same people!) have claimed that its value is 1. But we're talking about summation methods, not summation opinions. Every summation method I know (that's about 3 or so) gives $\frac{1}{2}$ for this series. – Pete L. Clark Mar 25 2010 at 3:34
As John Baez has pointed out, the people who claim its value is 0 and the people who claim its value is 1 should compromise and call it 1/2. (That's not quite as flippant as it sounds.) – Tom Leinster Mar 25 2010 at 8:23
@Tom While I agree that the sum is morally ${1\over 2}$, I'm tempted to call your bluff on the "That's not quite as flippant as it sounds" claim. :-) – Dan Piponi Mar 25 2010 at 18:00

Most summation methods come equipped with Tauberian theorems, which basically say that given some conditions on how quickly the terms diverge, then if the method gives an answer, that answer is basically unique. Most summations methods (that have stood the test of time) are neatly arranged in a hierarchy so that if a 'weaker' method works, then all 'stronger' methods (those which can deal with greater divergence) will work and give the same answer. Hardy's book covers all this material in detail.

Another good modern source is Balser's "From divergent series to analytic differential equations", which does a great job at digesting Ecalle's theory of resurgent functions and resommability and giving it back in terms that mere mortals can understand. You might also enjoy a nice overview, by Christiane Rousseau Divergent series: past, present, future.

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G. H. Hardy, DIVERGENT SERIES, page 73:

There is no general theorem for Abelian methods corresponding to Theorem 17: different methods may well sum the same series to different sums. Thus $1-1+1-\dots$ is summable $(\mathrm{A})$ to sum $\frac{1}{2}$, but summable $(\mathrm{A},\lambda)$, whern $(\lambda_n)$ is the sequence $0,1,3,4,6,7,\dots$, to $\frac{1}{3}$: see ยง 3.9.

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 Theorem 17 is that the Nörlund methods are all consistent. – Gerald Edgar Apr 28 2010 at 19:10