The Akra–Bazzi method handles recurrences of the form $$ f(n) = \sum_{l=1}^k a_l f(n/b_l). $$ Does it work when then number of $a \cdot f(n/b)$ is not finite, meaning that we have a sum that depends on our input $n$?
What about this:
$$ f(n) = \sum_{l=1}^{n} f(n/3^l) $$
Can I say that we need to find the $p$ so that
$$ \sum_{j=1}^{n} (\frac{1}{3^j})^p =1,$$
and because $n \rightarrow \infty$ we can say it is an infinite geometric sum? Or because $n$ is our input we can't say that, and the Akra–Bazzi method fails to deal with this kind of stuff? There is no internet site I found that says $k$ (which in Wikipedia is the number of $f(n/b)$) needs to be finite.