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 November 20th, 2013, 03:11 AM #1 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Summing Divergent Series This topic is dedicated to discuss about summability method and analysis of several divergent series. The applications and conditions for summability methods would be discussed, as well as deterministic approaches to understand the complicated relations of axioms relating summability methods and the converging ability of those techniques. For introductory purpose, we present some examples here : Consider the divergent series 1 - 2 + 3 - 4 + ... and 1 + 2 + 3 + 4 + ... Both are completely divergent in the Cauchy sense, although applying Abel summation to the former and Ramanujan summation to the later makes them both converge. If the Abel method is applied to the later sum, we see that the method positively fails and is unable to converge the sum. For another example, consider 1 + 1/2 + 1/3 + 1/4 + ... and 1 + 1 + 1 + 1 + ... both are some divergent forms of the p-series, namely, ?(0) and ?(-1), both taken in the sense that zeta here is the zeta of real part > 0, i.e., the mere p-series. Wee see that even if the later converges in C(0) Ramanujan summation, the later needs a more "powerful" C(1) summation. In both of these examples, we see that there are some divergent series which takes a fair bit of "hard" summation methods to converge. We formally describe the problem as thus : For some series S, S*, if S* takes more "stronger" summability methods to sum, then S* has stronger divergence than S. The formal definition for "strong" methods is : For some A and B, two summation methods with certain axioms of regularity, linearity and stability, or some weak forms of it, if A and B share consistency, i.e., for some series S that is both summable in A and B and A(S) = B(S), and if A sums mores series than B, then A is stronger than B. For the sake of acceleration methods, we one can use weaker conditions than consistency, without using the fact that they are equivalent at all, etc. I hope this is all for an introductory post, Balarka .
 November 20th, 2013, 03:17 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Summing Divergent Series As an addition, I post the conditions that (generally) identify a certain summability methods : Regularity : A summation method is regular if, whenever the sequence s converges to x, A(s) = x. Equivalently, the corresponding series-summation method evaluates A?(a) = x. Linearity : A is linear if it is a linear functional on the sequences where it is defined, so that A(kr + s) = kA(r) + A(s) for sequences r, s and a real or complex scalar k. Since the terms an = sn+1 ? sn of the series a are linear functionals on the sequence s and vice versa, this is equivalent to A? being a linear functional on the terms of the series. Stability : If s is a sequence starting from s(0) and s? is the sequence obtained by omitting the first value and subtracting it from the rest, so that s?(n) = s(n+1) ? s(0), then A(s) is defined if and only if A(s?) is defined, and A(s) = s(0) + A(s?). Equivalently, whenever a?(n) = a(n+1) for all n, then A?(a) = a(0) + A?(a?). Balarka .
 November 20th, 2013, 03:57 AM #3 Senior Member   Joined: Jan 2011 Posts: 120 Thanks: 2 Re: Summing Divergent Series Balarka, I don't know if this is relevant for your question, but I recently found a nice analytic continuation of $\zeta(s)$ that is containing the sum of two divergent series (see here: http://mathoverflow.net/questions/14...tion-for-zetas ). This is the formula that certainly works for $\Re(s) > 0$ but I believe also works for $\Re(s) > -1$. $\displaystyle \zeta(s)= \frac{1}{2\,(s-1)} \left(s+1+ \sum _{n=1}^{\infty } \left( {\frac {s-1-2\,n}{{n}^{s}}}+{\frac {s+1+2\,n}{\left( n+1 \right) ^{s}}}\right) \right)$ I like the symmetry of this formula and it also doesn't induce any additional zeros like the Dirichlet "alternating zeta" $\eta$-function does. Still very curious to understand more about these two divergent series that only seem to exactly balance out for certain $s$ when $\Re(s)=\frac12$ (assuming RH). Maybe your thread will provide more insights. Found (never posted) that the above could be reduced to an even simpler form with $\mathcal H(n)$ = Heaviside step function. Note $\mathcal H(0)=1$i.e. assumed to be right-continuous: $\zeta(s)= \sum _{n=-\infty }^{\infty } \dfrac{\frac12+{\frac {n+{\mathcal H} \left( n \right) }{s-1}}} {\left( \left| n \right| +{\mathcal H} \left( n \right) \right) ^{s}}$
November 20th, 2013, 04:35 AM   #4
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Re: Summing Divergent Series

Quote:
 Consider the divergent series 1 - 2 + 3 - 4 + ... and 1 + 2 + 3 + 4 + ... Both are completely divergent in the Cauchy sense, although applying Abel summation to the former and Ramanujan summation to the later makes them both converge
So this tells us something straight away. the transformation mapped onto the series 1 - 2 + 3 - 4 + ... by the Abel Summation is convergent and the transformation mapped onto the series 1 + 2 + 3 + 4 + ... is convergent.

They dont cause the series to be convergent, they perform transformations on the series as a result making them into another function altogether which does in fact converge.

November 20th, 2013, 05:31 AM   #5
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Re: Summing Divergent Series

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 Originally Posted by 667 They dont cause the series to be convergent, they perform transformations on the series as a result making them into another function altogether which does in fact converge.
Yes, speaking technically. I don't prefer calling them transformation or the like because it makes sound partial to Cauchy sense. They are, as the Cauchy definition is, definitions and spaces of convergence.

Quote:
 Originally Posted by Agno I don't know if this is relevant for your question, but I recently found a nice analytic continuation of zeta that is containing the sum of two divergent series
Hmm, quite nice. I never saw this expression before, but it seems that EM summation formula would be sufficient to prove it (iff my calculations are not incorrect).

The behavior of the divergent series are then the reason of the EM formula. Recall that there are certain (nontrivial) divergent series which sum up to a convergent series.

November 20th, 2013, 05:41 AM   #6
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Re: Summing Divergent Series

Quote:
 Recall that there are certain (nontrivial) divergent series which sum up to a convergent series.

This is in reference to two periodic series having a transformation function that has i constant finite value for all possible (x,y) that is a solution?

November 20th, 2013, 08:12 AM   #7
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Re: Summing Divergent Series

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 Originally Posted by 667 This is in reference to two periodic series having a transformation function that has i constant finite value for all possible (x,y) that is a solution?
This is referred to Agno's post, which concerns about an analytic continuation of zeta, this is what you are asking, I presume?

November 20th, 2013, 10:12 AM   #8
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Re: Summing Divergent Series

Quote:
Originally Posted by 667
Quote:
 Consider the divergent series 1 - 2 + 3 - 4 + ... and 1 + 2 + 3 + 4 + ... Both are completely divergent in the Cauchy sense, although applying Abel summation to the former and Ramanujan summation to the later makes them both converge
So this tells us something straight away. the transformation mapped onto the series 1 - 2 + 3 - 4 + ... by the Abel Summation is convergent and the transformation mapped onto the series 1 + 2 + 3 + 4 + ... is convergent.

They dont cause the series to be convergent, they perform transformations on the series as a result making them into another function altogether which does in fact converge.
Hmm... it's a matter of terminology. If you think of Cauchy-Weierstrass convergence as "the" definition, then all you're saying is that although these sums diverge they Abel-converge or Ramanujan-converge. But for a person who considers Abel convergence the 'right' definition, this example just shows that Cauchy-Weierstrass convergence is too weak of a concept.

In a universal sense the important thing to understand is how these definitions work together. In particular, if a sequence converges in the Cauchy-Weierstrass sense then not only does it converge in the Abel sense but it converges to the same value.

November 20th, 2013, 10:12 PM   #9
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Re: Summing Divergent Series

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 Originally Posted by CRGreathouse Hmm... it's a matter of terminology.
Exactly, as I have indicated a few posts back.

Quote:
 Originally Posted by CRGreathouse In particular, if a sequence converges in the Cauchy-Weierstrass sense then not only does it converge in the Abel sense but it converges to the same value.
Right, but no such "in general" identification of consistency exists for any two summability methods with Regularity and Stability. Actually, I might be wrong, but I think there doesn't exists one even with Linearity on the hand...

Balarka
.

 November 21st, 2013, 04:39 AM #10 Senior Member   Joined: Sep 2013 From: Perth, Western Australia Posts: 126 Thanks: 0 Re: Summing Divergent Series well from the way i see it, all of these methods are examples of ways we can classify series in terms of their convergence/ divergence. will post another example soon on my thread showing another way the method proposed can indicate what groups a series may can belong to in terms of its convergence/ divergence

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