This is a follow-up of this question.

It is known that the divergent 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. Analytic continuation of power series: 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 \dfrac{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.

  • $\begingroup$ Yes this is posible. $\endgroup$ – Gerben Mar 19 '16 at 21:20
  • $\begingroup$ @Gerben How so? $\endgroup$ – user76284 Jul 10 '19 at 4:54

As written the answer to all three is no, since

$$ \begin{align} 1.& \quad \lim_{x \to 1^-} \sum_{n=1}^{\infty} p_n x^n \geq \lim_{x \to 1^-} \sum_{n=1}^{m} p_n x^n = \sum_{n=1}^{m} p_n \text{ for all fixed $m$,} \\ 2.& \quad \lim_{x \to 0^+} \sum_{n=1}^\infty p_n n^{-nx} \geq \lim_{x \to 0^+} \sum_{n=1}^m p_n n^{-nx} = \sum_{n=1}^{m} p_n \text{ for all fixed $m$, and} \\ 3.& \quad \sum_{n=1}^\infty \dfrac{p_n}{n^s} \text{ diverges for $s \leq 2$.} \end{align} $$

You could consider the last one only a technicality, since it may still be possible to find an analytic continuation of the Dirichlet series for $s \neq 2$, where it appears to have a pole of order $2$. I'm not really familiar enough with these things to say.

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  • 1
    $\begingroup$ $\sum \frac{p_n}{n^s}$ diverges for $\operatorname{Re} s \leqslant 2$ even. I have no idea whether that has an analytic continuation to $0$ either. $\endgroup$ – Daniel Fischer Aug 29 '14 at 23:21
  • $\begingroup$ Whoops, right, thanks for the correction :) $\endgroup$ – Antonio Vargas Aug 29 '14 at 23:23

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