An elementary number theoretic infinite series - MathOverflow

An elementary number theoretic infinite series

2010-03-17T12:18:11Z

For a positive integer k, let d(k) be the number of divisors of k. So d(1)=1, d(p) =2 if p is a prime, d(6)=4, and d(12)=6.

What is the precise asymptotics of SUM_{k=1}^n 1/(kd(k))

Background:

1) This came up on the side in the polymath5 project.

2) There, Tim Gowers wrote: If nobody knows the answer, maybe that’s one for Mathoverflow, where I imagine a few minutes would be enough.

3) Asked: 14:17 Jerusalem time. (The first accurate answer: 17:44 Jerusalem time.)

4) Looking only at primes or only at integers with a typical number of divisors suggested a loglogn behavior, but looking at semiprimes indicates the sum is larger. I dont know how much larger.

5) I couldn't find an asnwer on the web. If there is an easy way searching for an answer that I missed this will be interesting too.

Follow up:

Great answers!! thanks. What about the sum

$\sum_{k=1}^n 1/(kd^2(k))$ ?

Answer by maks for An elementary number theoretic infinite series

2010-03-17T15:44:25Z

The correct asymptotic is $C \cdot (\log N)^{1/2}$. (c.f Selberg-Delange method).

Answer by David Hansen for An elementary number theoretic infinite series

2010-03-17T15:46:35Z

(edited)

The answer can be extracted from a paper of Ramanujan, "Some formulae in the analytic theory of numbers", no. 17 in his collected papers. There he gives, among other things, the formula

$\sum_{n\leq X} \frac{1}{d(n)} \sim \frac{X}{\sqrt{\log{X}}}\pi^{-\frac{1}{2}}\prod_{p}\sqrt{p^2-p}\log{\frac{p}{p-1}}$.

The answer to the original question can be extracted from this by partial summation.

As for the "follow-up", the answer is $\sum_{n \leq X} \frac{1}{n d(n)^2} \sim C (\log{X})^\frac{1}{4}$. Again, Selberg-Delange...

Answer by Victor Miller for An elementary number theoretic infinite series

2010-03-17T20:23:24Z

The idea (from the Selberg-Delange) method to doing this problem is the following steps:

1) Let $F(s) = \sum_{n\ge 1} \frac{1}{n^s d(n)} = \prod_{p} \left(1 + \sum_{k=1}^{\infty} \frac{1}{(k+1) p^{ks}} \right)$. The latter is by multiplicativity of $d(n)$.

2) If we look, instead at $G(s) = \prod_p \left( 1 + \frac{1}{2 p^s} \right)$ we can see that $F(s)/G(s)$ has a non-zero limit as $s \rightarrow 1$ from above. $G(s)$ corresponds in our original sum to restricting $n$ to be square-free.

3) $G(s)^2$ almost looks like $\zeta(s)$. Show that $G(s)^2/\zeta(s)$ also has a non-zero limit at $s \rightarrow 1$.

4) You then use some Tauberian theorems to show that since $H_n \sim \log n$ (which is the sum associated with $\zeta(s)$ then the corresponding sum for $G(s)$ (i.e. over the square-free $n$) is $\sim to \sqrt{\log n}$.