19
$\begingroup$

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,

  • 2
    You can if you approximate $p_n\rightarrow n\ln(n)$2011-11-22
  • 1
    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.2011-11-22
  • 0
    @anon: Yes, I am aware of that. I wrote that regularization and summability methods assigns finite *values* to infinite, divergent series.2011-11-22
  • 0
    @howdy :how? (text)2011-11-22
  • 0
    @anon: I'm sorry, I should have been more careful. I meant to say that $ 1+2+3+ \dots n "=" -1/12 $ when $ n \to \infty $.2011-11-22
  • 0
    @Max Use some of the methods here but replace n by nln(n): http://math.stackexchange.com/questions/39802/why-does-123-dots-1-over-122011-11-22
  • 0
    @howdy: please consider adding a more detailed descriptions of your ideas on this question as an answer :) .2011-11-22
  • 2
    @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.2011-11-22

1 Answers 1