How do you go about solving this recurrence? How do you make an estimation for the substitution method, when the recursion tree did not help so much?
I have a recurrence 
$$T(n) = 5\cdot T(n/3) + n (\log n)^2$$
And upon doing the recurrence tree, I got a rather difficult running time 
$$ \sum_{j=1}^{\log(n)/\log(3)}  \frac{5^{j-1} n\log_2^2[n/(3^{j-1})]}{(3^{j-1})^2} $$
How do you go about solving this recurrence? 
 A: There  is another  closely  related recurrence  that  admits an  exact
solution.  Suppose we  have  $T(0)=0$ and  $T(1)=T(2)=1$ and  for
$n\ge 3$
$$T(n) = 5 T(\lfloor n/3 \rfloor) + n \lfloor \log_3 n \rfloor^2.$$
It seems reasonable to use integer  values here as the running time of
an algorithm is a function of discrete quantities.
Furthermore let the base three representation of $n$ be
$$n = \sum_{k=0}^{\lfloor \log_3 n \rfloor} d_k 3^k.$$
Then  we can  unroll the  recurrence to  obtain the  following exact
formula for $n\ge 3$
$$T(n) = 5^{\lfloor \log_3 n \rfloor}
+ \sum_{j=0}^{\lfloor \log_3 n \rfloor -1} 
5^j \times (\lfloor \log_3 n \rfloor - j)^2 \times 
\sum_{k=j}^{\lfloor \log_3 n \rfloor} d_k 3^{k-j}.$$
Now to get  an upper bound consider a string of  value two digits to
obtain
$$T(n) \le 5^{\lfloor \log_3 n \rfloor}
+ \sum_{j=0}^{\lfloor \log_3 n \rfloor -1} 
5^j \times (\lfloor \log_3 n \rfloor - j)^2 \times 
\sum_{k=j}^{\lfloor \log_3 n \rfloor} 2 \times 3^{k-j}.$$
This simplifies to
$$\frac{1457}{32} 5^{\lfloor \log_3 n \rfloor}
-\frac{9}{2} \lfloor \log_3 n \rfloor^2 3^{\lfloor \log_3 n \rfloor}
-\frac{45}{2} \lfloor \log_3 n \rfloor 3^{\lfloor \log_3 n \rfloor}
- 45\times 3^{\lfloor \log_3 n \rfloor}
\\ + \frac{1}{4} \lfloor \log_3 n \rfloor^2
+ \frac{5}{8} \lfloor \log_3 n \rfloor
+ \frac{15}{32}.$$
This bound is actually attained and cannot be improved upon, just like
the lower bound,  which occurs with a one digit  followed by zeroes to
give
$$T(n) \ge 5^{\lfloor \log_3 n \rfloor}
+ \sum_{j=0}^{\lfloor \log_3 n \rfloor -1} 
5^j \times (\lfloor \log_3 n \rfloor - j)^2 \times 
3^{\lfloor \log_3 n \rfloor-j}.$$
This simplifies to
$$16\times 5^{\lfloor \log_3 n \rfloor}
- \frac{3}{2} \lfloor \log_3 n \rfloor^2 3^{\lfloor \log_3 n \rfloor}
- \frac{15}{2} \lfloor \log_3 n \rfloor 3^{\lfloor \log_3 n \rfloor}
- 15 \times 3^{\lfloor \log_3 n \rfloor}.$$
Joining the dominant terms of the upper and the lower bound we obtain
the asymptotics
$$5^{\lfloor \log_3 n \rfloor}
\in \Theta\left(3^{\log_3 5  \times \log_3 n}\right) 
= \Theta\left(n^{\log_3 5}\right).$$

This is in agreement with what the Master theorem would produce.
