An elementary number theoretic infinite series 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 $\sum_{k=1}^n 1/(kd^2(k))$ ?
 A: 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 $\sim to \sqrt{\log n}$.
A: The correct asymptotic is $C \cdot (\log N)^{1/2}$. (c.f Selberg-Delange method).
A: (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...
