- #1

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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.