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 are the precise asymptotics of $\sum_{k=1}^n 1/(k d(k))$?

## 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 $\log\log n$ behavior, but looking at semiprimes indicates the sum is larger. I don't know how much larger.

5) I couldn't find an answer 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, $d(p) = 2$. $\endgroup$ – Victor Miller Mar 18 '10 at 10:38