My opinion is ; 


We may use id(d)=d arithmetic function and log\*id dirichlet convolution in the question.
i thought that ; when we multiply and divide n with $(\log d) / d$ we obtain 
$F(S)=\sum\_{n=1}^{\infty} \frac{1}{n^{s+1}} \cdot\left(\log\*id\right)$
so 


$F(S)=D(log,s+1).D(id,s+1)$


So we get


$F(S)=-\zeta^{\prime}(s+1). \zeta(s)$


i am curious that my solution is right or not.You may write your solutions and different ideas. thanks for your helps.



What is the dirichlet series of $f(n)=\sum_{d | n}(\log d) / d$ function?
We can verify this by direct calculation. By definition,
$$F(s)=\sum\_{n=1}^\infty\frac{f(n)}{n^s}=\sum\_{n=1}^\infty\frac{1}{n^s}\sum\_{d\mid n}\frac{\log d}{d}.$$
We rearrange the double sum so that $d$ comes first, and then write $n=dm$ to separate variables. We get
\begin{align\*}
F(s)&=\sum\_{d=1}^\infty\frac{\log d}{d}\sum\_{m=1}^\infty\frac{1}{(dm)^s}=\left(\sum\_{d=1}^\infty\frac{\log d}{d^{s+1}}\right)\left(\sum\_{m=1}^\infty\frac{1}{m^s}\right).
\end{align\*}
On the right hand side, the first factor is $-\zeta'(s+1)$, while the second factor is $\zeta(s)$, so
$$F(s)=-\zeta'(s+1)\zeta(s).$$


**P.S.** Your question is not of research level, but I felt like answering it.


