Solving recursion $T(n)=2T(n/2) +2/\log(n)$
Writing the recursion, we have:

$$\begin{align}
T(n)&amp;=2T(n/2)+2/\lg n\\&amp;
= 2(2T(n/4)+2/\lg(n/2))+2/\lg n\\&amp;
= 4T(n/4)+4/(\lg n-1)+2/\lg n\\&amp;
= 4(2T(n/8)+2/\lg(n/4))+4/(\lg n-1)+2/\lg n\\&amp;
= 8T(n/8)+8/(\lg n - 2)+4/(\lg n-1)+2/\lg n\\&amp;
\;\,\vdots\\&amp;
= 2^kT(n/2^k)+\sum_{i=1}^k\frac{2^i}{\lg n - i + 1},\\&amp;
\end{align}
$$
replacing $k$ with $\lg n$, we found:

$$T(n)=nT(1)+\sum_{i=1}^{\lg n}\frac{2^i}{\lg n - i + 1}.$$

As mentionned in the comments, by choosing $j=\lg n - i + 1$ in the summation ($i = \lg n - j + 1$), we obtain:

$$\begin{align}T(n)&amp;=nT(1)+\sum_{j=\lg n}^{1}\frac{2^{\lg n - j + 1}}{j}\\&amp;=nT(1)+\sum_{j=1}^{\lg n}\frac{2n}{j2^j}\\&amp;=nT(1)+2n\sum_{j=1}^{\lg n}\frac{1}{j2^j}\\&amp;=n\Bigl(T(1)+2\sum_{j=1}^{\lg n}\frac{1}{j2^j}\Bigr).\end{align}$$ 

We finally obtain $$T(n)=\Theta(n).$$