23
$\begingroup$

It is possible, by means of zeta function regularization and the Ramanujan summation method, to assign a finite value to the sum of the natural numbers (here $n \to \infty $) :

$$ 1 + 2 + 3 + 4 + \cdots + n \; {“ \;=\; ”} - \frac{1}{12} . $$

Is it also possible to assign a value to the sum of primes, $$ 2 + 3 + 5 + 7 + 11 + \cdots + p_{n} $$ ($n \to \infty$) by using any summation method for divergent series?

This question is inspired by a question on quora.

Thanks in advance,

$\endgroup$
8
  • 3
    $\begingroup$ You can if you approximate $p_n\rightarrow n\ln(n)$ $\endgroup$
    – TROLLHUNTER
    Nov 22, 2011 at 18:08
  • 2
    $\begingroup$ If you can, it won't be as nice or have as much meaning as it does for the zeta function. Your first sum can be more concretely written as $\zeta(-1)=\frac{1}{12}$. However, the prime zeta function cannot be analytically continued to the left of the imaginary axis. $\endgroup$
    – Eric Naslund
    Nov 22, 2011 at 18:18
  • $\begingroup$ @anon: Yes, I am aware of that. I wrote that regularization and summability methods assigns finite values to infinite, divergent series. $\endgroup$
    – Max Muller
    Nov 22, 2011 at 18:26
  • $\begingroup$ @howdy :how? (text) $\endgroup$
    – Max Muller
    Nov 22, 2011 at 18:33
  • 3
    $\begingroup$ @Max: Note that $p_n\sim n\log n$ by the prime number theorem. You can zeta-regularize the divergent sum $\sum_{n=1}^\infty n\log n$ by evaluating $-\zeta'(-1)=\log A-1/12$, where $A$ is the Glaisher-Kinkelin constant. So it's an answer to something similar to your question. $\endgroup$
    – anon
    Nov 22, 2011 at 18:48

2 Answers 2

Reset to default
28
$\begingroup$

Fröberg shows in his paper that the prime zeta function

$$P(s)=\sum_{p\in \mathbb P} \frac1{p^s}=\sum_{k=1}^\infty \frac{\mu(k)}{k}\log\zeta(ks)$$

where $\mu(k)$ and $\zeta(s)$ are respectively the Möbius and Riemann functions, cannot be analytically continued to the left half-plane, $\Re\,s\leq 0$ (in particular, we cannot give a reasonable evaluation of $P(-1)$), due to the clustering of poles along the imaginary axis arising from the nontrivial zeros of the Riemann $\zeta$ function.

prime zeta plots

Note the nasty-looking left edges in both plots above.

This result is originally due to Landau and Walfisz. See the linked papers for more details.

$\endgroup$
5
  • 3
    $\begingroup$ Notes: The formula is basically the infinite version of Mobius inversion. And the $P(-1)$ can't be defined because a dense line of poles forms a blockade against any hope of analytic continuation. $\endgroup$
    – anon
    Nov 22, 2011 at 18:20
  • 2
    $\begingroup$ Ok, but perhaps there is a different method by means of which it can be done, right? Analytic continuation is only one of many ways to sum divergent series. $\endgroup$
    – Max Muller
    Nov 22, 2011 at 18:22
  • 1
    $\begingroup$ @J.M.: Here is a short proof: In your formula above, notice that when $s=\frac{1}{k}$ for some squarefree $k$, then we have a $\log\zeta(1)$ term appear in the sum, which means it is a singular point. This sequence $\frac{1}{k}$ for $k$ squarefree is then a sequence of singularities which converges to zero. $\endgroup$
    – Eric Naslund
    Nov 22, 2011 at 18:28
  • 1
    $\begingroup$ Also note that this result is originally due to Landau and Walfisz. $\endgroup$
    – Eric Naslund
    Nov 22, 2011 at 18:31
  • $\begingroup$ Thanks, Eric and anon, for the added details. $\endgroup$
    – J. M. ain't a mathematician
    Nov 22, 2011 at 18:39
2
$\begingroup$

So I received a comment about this on this question. So if its possible to sum the primes using the techniques of that question one strategy would be to extract the $x^1$ term of $\sum_{n=0}^{\infty} e^{p_n x}$ or the $-\frac{1}{x}$ term from the asymptotic expansion of

$$ \sum_{n=0}^{\infty} e^{-\frac{p_n}{x}}$$

enter image description here

So we have that

$$ \sum_{n=0}^{\infty} e^{-\frac{p_n}{x}} = \pi(x) + \frac{1}{2} + ... $$

Where the $\pi(x)$ is a continuous version of the prime counting function (however one can compute that) and $...$ is currently unknown. This result is from a conjectured asymptotic limit. The problem with this formula is that it requires us to define $p_0$ which I don't believe we have a definition for at this time.

I wish had better techniques (or at least A non guess and check technique) for computing these asymptotic series. I think that such numerical techniques exist I just don't know what they are. User 2734364041 outlines a way here to use residue theorems and the mellin transform to get 'closed' forms for these types of sums but I am concerned slightly that using the prime function won't work because of natural boundaries (his Mellin transform ends up yielding a prime zeta function which already one of the answerers here has pointed out sucks)

$\endgroup$
1
  • 1
    $\begingroup$ @Siddharth Thank you! Perhaps one can use the estimate $p_n \sim n \log n$ for your method? Might be interesting to confirm whether your method yields the same answer as anon's, as provided in a comment to the question (above) $\endgroup$
    – Max Muller
    Jun 18 at 20:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .