Big-O Solving Recurrence Relation by iteration with fractions
We can rewrite your recursion as
$$
B(n) = B(n-4) + \Theta\left(\frac{1}{n}\right).
$$
It follows that
$$
\begin{align*}
B(n) &amp;= \Theta\left(\frac{1}{n} + \frac{1}{n-4} + \frac{1}{n-8} + \cdots\right) \\ &amp;=
\Theta\left(\frac{1}{n/4} + \frac{1}{n/4-1} + \frac{1}{n/4-2} + \cdots \right) \\ &amp;=
\Theta(\log (n/4)) = \Theta(\log n),
\end{align*}
$$
using the formula for the sum of a harmonic series.