Summation methods for divergent series - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T07:17:16Z http://mathoverflow.net/feeds/question/19201 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19201/summation-methods-for-divergent-series Summation methods for divergent series Eric O. Korman 2010-03-24T15:16:49Z 2012-11-20T06:52:27Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/19201/summation-methods-for-divergent-series/19211#19211 Answer by David Speyer for Summation methods for divergent series David Speyer 2010-03-24T16:47:29Z 2012-11-20T06:52:27Z <p>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 <em>Divergent Series</em>; the proof is short enough to give here.</p> <p>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 <code>$$c_n = (n+1)(L + o(1)).$$</code></p> <p>Let $A(x) = \sum a_i x^i$. Then $\sum c_i x^i = A(x)/(1-x)^2$ so <code>$$A(x) = (1-x)^2 \sum_{n=0}^{\infty} (L (n+1) + o(n+1)) x^n$$</code> where the $o$ is as $n \to \infty$, independent of $x$. But <code>$$\sum_{n=0}^{\infty} (n+1) x^n = 1/(1-x)^2$$</code> so <code>$$A(x) = L (1-x)^2/(1-x)^2 + o \left( (1-x)^2/(1-x)^2\right) = L+ o(1) \quad \mbox{as} \ x \to 1^{-}.$$</code> So the Abel sum of $a_i$ is also $L$.</p> <hr> <p>Deleted an argument that Cesaro summability implies zeta summability; not sure I can sum by parts where needed.</p> <hr> <p>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.</p> http://mathoverflow.net/questions/19201/summation-methods-for-divergent-series/19217#19217 Answer by Dan Piponi for Summation methods for divergent series Dan Piponi 2010-03-24T17:45:02Z 2010-03-24T17:45:02Z <p>The series 1-1+1-1+... has a <a href="http://en.wikipedia.org/wiki/History_of_Grandi%27s_series" rel="nofollow">long history</a> of people assigning it different values using different methods.</p> http://mathoverflow.net/questions/19201/summation-methods-for-divergent-series/19251#19251 Answer by Jacques Carette for Summation methods for divergent series Jacques Carette 2010-03-25T00:01:42Z 2010-04-28T23:30:19Z <p>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.</p> <p>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 <a href="http://www.dms.umontreal.ca/~rousseac/divergent.pdf" rel="nofollow">Divergent series: past, present, future</a>.</p> http://mathoverflow.net/questions/19201/summation-methods-for-divergent-series/22889#22889 Answer by Gerald Edgar for Summation methods for divergent series Gerald Edgar 2010-04-28T19:07:37Z 2010-04-28T19:07:37Z <p>G. H. Hardy, DIVERGENT SERIES, page 73:</p> <p>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 § 3.9.</p>