Applying Akra–Bazzi with an unbounded number of summands
Quoting from Wikipedia, the Akra–Bazzi method applies to recurrences of the form
$$
T(x) = g(x) + \sum_{i=1}^k a_i T(b_i x + h_i(x))
$$
where $a_i,b_i$ are constant, $a_i&gt;0$, $0&lt;b_i&lt;1$, $g$ is polynomially bounded, and $h$ is &quot;sublinear enough&quot;.
Your recurrence is not of this form, so you cannot apply the Akra–Bazzi method. It could be that the techniques used to prove the Akra–Bazzi theorem also apply to your case, but you will have to look at the proof.

Notice that
$$
f(n) - f(n/3) = \sum_{i=1}^{\log_3 n} f(n/3^i) - \sum_{i=1}^{\log_3 (n/3)} f(n/3^{i+1}) = f(n/3),
$$
and so $f$ satisfies the recurrence
$$
f(n) = 2f(n/3).
$$
In particular,
$$
f(n) = 2^{\log_3 n} f(1) = n^{\log_3 2} f(1).
$$
(Throughout, we are assuming that $n = 3^k$.)