Trying to find a substitution to solve a recurrence
You don't need to try that hard. Instead, you can notice that on the one hand clearly $B(n) = \Omega(n)$, and on the other hand
$$
\begin{align*}
B(n) &amp;= \Theta\left(n + \frac{n}{\log n} + \frac{n}{\log n \log(n/\log n)} + \cdots\right) \\ &amp;=
O\left(n + \frac{n}{2} + \frac{n}{4} + \cdots\right) = O(n).
\end{align*}
$$
The first inequality holds since for $n$ larger than some constant, $\log_2 n \geq 2$ (you just need to stop the recursion once $n$ gets too small, at the cost of additional $O(1)$ which gets swallowed in the big Theta). The second inequality holds since $n+n/2+n/4+\cdots = 2n$.

Formally speaking, you can argue that $B(n) = B(n/2) + \Theta(n)$ grows faster than $B(n) = B(n/\log_2 n) + \Theta(n)$.