Master Theorem applicable here? Let
$T(n):=\begin{cases} \frac{2+\log n}{1+\text{log}n}t(\lfloor\frac{n}{2}\rfloor) + \log ((n!)^{\log n}) &amp; \text{if }n&gt;1 \\
1 &amp; \text{if }n=1
\end{cases}$

I need to prove that $t(n) \in O(n²)$, thus $t(n) \leq c\cdot n²$

I asked the question here and I got really great help last time, the thing is after I was shown last time that $f(n)=\log(n)\cdot\log(n!)$ is $\Theta(2\cdot\log(n)\cdot n) = \Theta(\log(n)\cdot n)$ I thought I could then use the master theorem

However since $a=\frac{2+\log n}{1+\log n}$ is NOT a constant I can not use the master theorem
but I thought that I could use an upper bound for $a$, since $\frac{2+\log n}{1+\log n} &lt; 2 \quad\forall n$ and then use the master theorem for $a=2$, $b=2$. But am I allowed to use the master theorem after finding an upper bound for the non-constant $a$? 

What would other ways be to show that $T(n) = O(n^2)$ ?
Yes, you can define $T'(n) = 2 T'(\frac{n}{2}) + \Theta(n \log n)$, notice that $T(n) \le T'(n)$, and use the master theorem on $T'$ to obtain an upper bound of $O(n \log^2 n)$ to $T$.

Since for $n \ge 2$, $\frac{2+\log n}{1+\log n} \le \frac{3}{2}$ you can get a better upper bound by comparing $T$ to
$$T'' = \begin{cases}\frac{3}{2}T''(\frac{n}{2}) + \Theta(n \log n) &amp; \text{if $n\ge2$} \\ \Theta(1) &amp; \text{otherwise}\end{cases}.$$ Applying the master theorem on $T''$ yields and upper bound of $O(n \log n)$ for $T$.