I know the result of $T(n)=2T(\frac{n}{2})+\frac{n}{\log n}$=$\Theta(n\log(\log(n)))$,
But I am lost at
$Here,\qquad T(n)=2T(\frac{n}{2})+\frac{n}{\log n}$
$T(\frac{n}{2})=2T(\frac{n}{4})+\frac{n}{2\log\frac{n}{2}}$
$T(\frac{n}{4})=2T(\frac{n}{8})+\frac{n}{4\log\frac{n}{4}}$
$T(n)=2[2T(\frac{n}{4})+\frac{n}{2\log\frac{n}{2}}]+\frac{n}{\log n}$
$T(n)=4T(\frac{n}{4})+\frac{n}{\log\frac{n}{2}}+\frac{n}{\log n}$
$T(n)=4[2T(\frac{n}{8})+\frac{n}{\log\frac{n}{4}}]+\frac{n}{\log\frac{n}{2}}+\frac{n}{\log n}$
$T(n)=8T(\frac{n}{8})+\frac{n}{\log\frac{n}{4}}+\frac{n}{\log\frac{n}{2}}+\frac{n}{\log n}$
.
.
.
$T(n)=2^kT(\frac{n}{2^k})+\frac{n}{\log\frac{n}{2^{k-1}}}+\ldots+\frac{n}{\log\frac{n}{4}}+\frac{n}{\log\frac{n}{2}}+\frac{n}{\log\frac{n}{2^0}}$
I know $$\sum_{k = 1}^n \frac{1}{i} = \Theta(\log(n))$$
where might I be missing to bring my result in the form of harmonic series?
