Leaving recurrence summation in terms of $k$, $\sum_{i=0}^{k-1}\frac{3^i\sqrt{\frac n{3^i}}}{\log\frac n{3^i}}$ I have an exercise where I need to use the substitution method to solve the following recurrence and determine their corresponding complexity.
$$t(n)=3t(n/3) + \frac{\sqrt n}{\log n}$$
After some iterations, I got the following pattern.
$$t(n)=t\left(\frac{n}{3^k}\right)+\sum_{i=0}^{k-1}\frac{3^i\sqrt{n/3^i}}{\log n/3^i}$$
Honestly I do not know what kind of approach I could use to solve the summation, and leave everything in terms of $k$.

Hint 1. As regards the asymptotic analysis (complexity?), you may use the Master Theorem. Then $c_{crit}=\log_2(3)\approx 1.58$ and $f(n)=\frac{n^{1/2}}{\log(n)}\leq n^{1/2}$. What about $T(n)$?
Hint 2. Note that
$$t(2^n)=3^nt(1)+\sum_{k=0}^{n-1}\frac{3^k\sqrt{2^{n-k}}}{\log (2^{n-k})}
=3^nt(1)+3^n\sum_{k=1}^{n}\frac{(\sqrt{2}/3)^k}{k\log (2)}\sim C\cdot 3^n$$
because the series $\sum_{k=1}^{n}\frac{(\sqrt{2}/3)^k}{k\log (2)}$ is convergent (note that $\sqrt{2}/3<1$).
