Big-O Solving Recurrence Relation by iteration with fractions I was trying to solve the recurrence relation in order to get a some big-O bound $$ B(n) = B(n-4) + \frac{1}{n} + \frac{5}{n^{2} + 6} + \frac{7n^{2}}{3n^{3} + 8}$$  by following the accepted answer here. I expanded which resulted in $$ B(n) = B(n-8) + \frac{1}{n-1} + \frac{5}{(n-1)^{2} + 6} + \frac{7(n-1)^{2}}{3(n-1)^{3} + 8} + \frac{1}{n} + \frac{5}{n^{2} + 6} +  \frac{7n^{2}}{3n^{3} + 8}$$

This lead to $$ B(n) = B(n-4k) + \frac{1}{n-k} + \frac{5}{(n-k)^{2} + 6} + \frac{7(n-k)^{2}}{3(n-k)^{3} + 8} + \frac{1}{n-k} + \frac{5}{(n-k)^{2} + 6} +  \frac{7(n-k)^{2}}{3(n-k)^{3} + 8}$$ However, I get stuck when replacing $k$ with $n$ since this results in dividing by zero. How do I proceed? Can it solved by the iteration?
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.