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:
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1) This came up on the side in the [polymath5](http://gowers.wordpress.com/2010/03/13/edp12-representing-diagonal-maps/#comment-6697) 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:
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Great answers! Thanks. What about the sum $\sum\_{k=1}^n 1/(kd^2(k))$ ?



An elementary number theoretic infinite series
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 $\sim to \sqrt{\log n}$.


