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Convergence of a sum over primes

  1. Sep 29, 2015 #1
    I am trying to understand a condition for a nonincreasing sequence to converge when summed over its prime indices. The claim is that, given [itex]a_n[/itex] a nonincreasing sequence of positive numbers,
    then [itex]\sum_{p}a_p[/itex] converges if and only if [itex]\sum_{n=2}^{\infty}\frac{a_n}{\log(n)}[/itex] converges.

    I have tried various methods to prove this but my error estimates are always too large.
    The closest I came to a proof is this: first, extend the sequence [itex]a_n = a(n)[/itex] to positive reals by "connecting the dots" (interpolating by some nondecreasing function that takes on the same values as [itex]a_n[/itex] on the integers. Then, do the same for [itex]\pi(x)[/itex] (prime counting function) and [itex]p(x)[/itex] (the n-th prime). The goal is to use the integral test to relate the two sums.

    So [itex]\sum_{p}a_p[/itex] converges if and only if [itex]\int_{1}^{\infty}a(p(x))dx[/itex] converges.
    Using the substitution [itex] t = p(x) [/itex] (so [itex]x = \pi(t), dx = \pi'(t)dt[/itex]), the second integral equals [itex]\int_{2}^{\infty}a(t)\pi'(t)dt[/itex]. By the Prime Number Theorem, [itex]\pi(t) = \frac{t}{\log(t)} + O\left(\frac{t}{\log^2(t)}\right)[/itex], so the derivative is (intuitively) [itex]\pi'(t) = \frac{1}{\log(t)} + O\left(\frac{1}{\log^2(t)}\right)[/itex]. So, assuming this "intuition" is correct, the integral is [itex]\int_{2}^{\infty}\frac{a(t)}{\log(t)}dt + ...[/itex] where the ellipsis are terms of lower order than the main term. This integral converges if and only if [itex]\sum_{n=2}^{\infty}\frac{a_n}{\log(n)}[/itex] converges.

    That would be good, but I am unable to prove the "intuitive" step. I need some estimate on the order of the derivative of [itex]\pi(x)[/itex], but the only information I have is the big-Oh of the function, not its derivative.
  2. jcsd
  3. Sep 29, 2015 #2
    can pi(x) be defined for non integer x as the linear interpolation between neighboring integer values? then the derivative has some meaning and the proof looks pretty convincing
  4. Sep 29, 2015 #3
    Well, that's what I want to do. My problem is how to rigorously prove that the slopes of the linear interpolation grow with order 1/log(x). The main issue is that pi(x) will be constant most of the time, except at the primes where it jumps by 1, so there will be infinitely many points where the slope will be 1, so it can't be O(1/log(x)).
  5. Sep 29, 2015 #4
    pi(x) , meaning the linearly interpolated version, has a slope of 1/distance between neighboring primes, doesn't it?
  6. Sep 29, 2015 #5
    Oh oops. Yeah it would be 1 / (p(n+1) - p(n)). But still, assuming the twin prime conjecture is true, there would be infinitely many cases where the slope is 1/2... Clearly this happens rarely, but does it compensate for the desired O(1/log x) order?
    Last edited: Sep 29, 2015
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