- #1
Boorglar
- 210
- 10
I am trying to understand a condition for a nonincreasing sequence to converge when summed over its prime indices. The claim is that, given a_n a nonincreasing sequence of positive numbers,
then \sum_{p}a_p converges if and only if \sum_{n=2}^{\infty}\frac{a_n}{\log(n)} 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 a_n = a(n) to positive reals by "connecting the dots" (interpolating by some nondecreasing function that takes on the same values as a_n on the integers. Then, do the same for \pi(x) (prime counting function) and p(x) (the n-th prime). The goal is to use the integral test to relate the two sums.
So \sum_{p}a_p converges if and only if \int_{1}^{\infty}a(p(x))dx converges.
Using the substitution t = p(x) (so x = \pi(t), dx = \pi'(t)dt), the second integral equals \int_{2}^{\infty}a(t)\pi'(t)dt. By the Prime Number Theorem, \pi(t) = \frac{t}{\log(t)} + O\left(\frac{t}{\log^2(t)}\right), so the derivative is (intuitively) \pi'(t) = \frac{1}{\log(t)} + O\left(\frac{1}{\log^2(t)}\right). So, assuming this "intuition" is correct, the integral is \int_{2}^{\infty}\frac{a(t)}{\log(t)}dt + ... where the ellipsis are terms of lower order than the main term. This integral converges if and only if \sum_{n=2}^{\infty}\frac{a_n}{\log(n)} 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 \pi(x), but the only information I have is the big-Oh of the function, not its derivative.
then \sum_{p}a_p converges if and only if \sum_{n=2}^{\infty}\frac{a_n}{\log(n)} 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 a_n = a(n) to positive reals by "connecting the dots" (interpolating by some nondecreasing function that takes on the same values as a_n on the integers. Then, do the same for \pi(x) (prime counting function) and p(x) (the n-th prime). The goal is to use the integral test to relate the two sums.
So \sum_{p}a_p converges if and only if \int_{1}^{\infty}a(p(x))dx converges.
Using the substitution t = p(x) (so x = \pi(t), dx = \pi'(t)dt), the second integral equals \int_{2}^{\infty}a(t)\pi'(t)dt. By the Prime Number Theorem, \pi(t) = \frac{t}{\log(t)} + O\left(\frac{t}{\log^2(t)}\right), so the derivative is (intuitively) \pi'(t) = \frac{1}{\log(t)} + O\left(\frac{1}{\log^2(t)}\right). So, assuming this "intuition" is correct, the integral is \int_{2}^{\infty}\frac{a(t)}{\log(t)}dt + ... where the ellipsis are terms of lower order than the main term. This integral converges if and only if \sum_{n=2}^{\infty}\frac{a_n}{\log(n)} 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 \pi(x), but the only information I have is the big-Oh of the function, not its derivative.