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

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

`$\sum_n 1/(n d(n)^2)$`

should be $\sim C (\log x)^{1/4}$, since if $p$ is prime, $d(p) = 2$. – Victor Miller Mar 18 '10 at 10:38