It is known that the series $$ P := \sum_{n=1}^{\infty} p_{n} \qquad \text{where } p_{n} \text{ is the n'th prime} $$ cannot be summed by means of (prime) zeta function regularization. (The result was originally due to Landau and Walfisz, see this paper. Froberg later showed it as well.)

However, there are loads of other summation methods. I am wondering whether any of the following summation methods can sum the divergent series of primes. For example:

  1. Abel summation/analytic continuation of power series (what is the difference?): Does $\lim_{x \to 1^{-} } \sum_{n=1}^{\infty} p_{n} x^{n} $ exist?
  2. Lindelöf summation: Does $\lim_{x \to 0} \sum_{n=1}^{\infty} p_{n} n^{-nx} $ exist?
  3. Analytic continuation of Dirichlet series: Does $\lim_{s \to 0} \sum_{n=1}^{\infty} \frac{p_{n}}{n^{s}} $ exist?

Do any of these methods or another summation method for assigning a number to the sum of primes work? If so, please also indicate what the closed form of the corresponding function (for which the limit exists) is.

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    $\begingroup$ The limits in (1) and (2) are infinite by straightforward comparisons; I don't know what you mean by (3) since the series does not converge for any value of $s$ near 0. $\endgroup$ – Mike Jury Aug 27 '14 at 12:48
  • $\begingroup$ I think (3) is the best bet for anything to work. Once a functional equation is established for the Dirichlet series I would imagine that routine-ish techniques would take over from there. The functional equation would be the tricky part though. $\endgroup$ – BSteinhurst Aug 27 '14 at 12:57
  • $\begingroup$ @MaxMuller (off-topic) I happened to notice that your user page links to an earlier user page. If you need your accounts merged, that can be done by filling out the form here mathoverflow.net/contact $\endgroup$ – j.c. Aug 27 '14 at 13:14
  • $\begingroup$ @j.c. Ah ok, thank you for the tip! $\endgroup$ – Max Muller Aug 27 '14 at 13:21
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    $\begingroup$ There is a difference between Abel summation and analytic continuation of power series. Abel summation requires that the power series is already analytic in the unit disc. $\endgroup$ – GH from MO Aug 27 '14 at 14:37

This question is poorly formulated, so I am not sure if it can be properly answered. I have some comments, rather then an answer. First, the methods 1,2 (Abel and Lindelöf summation), as well as any other linear regular summation method with positive matrix, are useless when dealing with a series with positive summands: if the original series is not convergent, then the summation does not help. Note that almost all of the standard methods are of this kind.

Analytic continuation methods, including zeta function regularization, is another story. But this is fishy: even if the continuation exists, it may well depend on the path. Besides, which function to continuate? I am at the opinion that a question like "Do any of these methods work?", in the context of analytic continuation, simply does not make any sense. What makes sense is "Does THIS method work?", and the choice of the method should be well motivated. Of course, motivation here may depend on the purpose of calculating the sum of a series, not only on the series itself: for example, zeta function regularization is a reasonable choice for a calculation of the Casimir force, because it can be shown to produce a correct physical result.

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    $\begingroup$ Just with regards to the last point, I would say that zeta function regularization works because it is giving the constant term when the divergence is smoothly cut off. I never learned this in my physics courses, though, and when I first read about it on Terry Tao's blog it was a revelation. $\endgroup$ – Aaron Bergman Aug 27 '14 at 20:17
  • $\begingroup$ "any other linear regular summation method with positive matrix, are useless when dealing with a series with positive summands: if the original series is not convergent, then the summation does not help." Do you have a reference for this? I also thought analytic continuation was unique. $\endgroup$ – user76284 Feb 25 at 19:10
  • $\begingroup$ user76284 @: The analytic continuation of a given function is (more or less) unique, but in this context we not not HAVE a function. We have to chose it and the number of choices is pretty much unlimited. For a reference to the above, see e.g. Hardy, Divergent series, Theorem 9 (p. 52 in the edition I am looking at). $\endgroup$ – Alex Gavrilov Feb 26 at 14:39

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