Understanding an Approximation I am reading the paper A Group-theoretic Approach to Fast Matrix Multiplication and there is an approximation in the paper I don't fully understand.
In the proof of Theorem 3.3. it is stated that
$$
 \frac{\ln (n(n+1)/2)!)}{\ln (1!2!\dots n!)}= 2+ \frac{2-\ln 2}{\ln n}+O\left( \frac{1}{\ln^2 n}\right).
$$
This is how far I got:
I used Stirling's Approximation to estimate the numerator, leading to
$$
\ln \left( \frac{n(n+1)}{2}!\right)=  \frac{n(n+1)}{2} \ln \left( \frac{n(n+1)}{2}\right)- \frac{n(n+1)}{2}+O\left(\ln \frac{n(n+1)}{2}\right).
$$
For the denominator I also use Stirling's approximation and $\ln (ab) = \ln a + \ln b$, leading to
$$
\ln (1!2!\dots n!) = \sum_{k=1}^n k\ln k -k + O(\ln k).
$$
I couldn't find a nice closed-form expression for $\sum_{k=1}^n k\ln k$ (is there one?),
so I approximated it by $\int x \ln x \, dx =  \frac{x^2\ln x} 2 -\frac {x^2} 4$, leading to
$$
 \ln (1!2!\dots n!)  \approx \frac{n^2 \ln n} 2 - \frac{3n^2 } 4 +O(n\ln n).
$$ 
Now, since the leading term of our fraction is, after cancelling, about $\frac{\ln(n^2)}{\ln n}$, I can see where the leading $2$ on the RHS comes from. But what about $\frac{2-\ln 2}{\ln n}$? Can it be calculated by a more precise bound on the denominator, e.g. by avoiding the integral?

The highest order term in $\log \prod\limits_{k=1}^n k!$ is $\frac{n^2}{2}\log n$, so in the desired
$$\log \left(\frac{n(n+1)}{2}\right){\Large !} = \left(2 + \frac{2-\log 2}{\log n} + O\left(\frac{1}{\log^2 n}\right)\right)\log \prod_{k=1}^n k!,\tag{1}$$
we get on the right hand side a term $O\left(\frac{n^2}{\log n}\right)$ which we cannot (need not) specify more precisely. Hence in the approximations, we can ignore every term of order $\frac{n^2}{\log n}$ or less.
What we need to establish $(1)$ is
$$\begin{align}
\log \left(\frac{n(n+1)}{2}\right){\Large !} &= \frac{n^2}{2}\log \frac{n(n+1)}{2} - \frac{n^2}{2} + O(n\log n),\tag{2}\\
\log \prod_{k=1}^n k! &= \frac{n^2}{2}\log n - \frac{3n^2}{4} + O(n\log n).\tag{3}
\end{align}$$
We obtain
$$\begin{align}
\log \left(\frac{n(n+1)}{2}\right){\Large !} - 2\log \prod_{k=1}^n k! &=
\frac{n^2}{2} \log n(n+1) - \frac{n^2}{2}\log 2 - \frac{n^2}{2}\\
&\quad - \frac{n^2}{2} \log n^2 + \frac{3n^2}{2} + O(n\log n)\\
&= \frac{n^2}{2}\log\frac{n+1}{n} + \frac{n^2}{2}(2-\log 2) + O(n\log n)\\
&= \frac{n^2}{2}(2-\log 2) + O(n\log n)\\
&= \frac{2-\log 2}{\log n}\left(\log \prod_{k=1}^nk! + O(n^2)\right) + O(n\log n)
\end{align}$$
and hence
$$\log \left(\frac{n(n+1)}{2}\right){\Large !} = \left(2 + \frac{2-\log 2}{\log n}\right)\log \prod_{k=1}^n k! + O\left(\frac{n^2}{\log n}\right).$$
Since $\log \prod k! \in \Theta\left(n^2\log n\right)$, the division yields
$$\frac{\log \left(\frac{n(n+1)}{2}\right){\Large !}}{\log \prod\limits_{k=1}^nk!} = 2 + \frac{2-\log 2}{\log n} + O\left(\frac{1}{\log^2 n}\right)$$
as desired.
It remains to establish $(2)$ and $(3)$. For that, we use the most significant terms of Stirling's approximation
$$\log k! = \left(k+\frac12\right)\log k - k + \frac12\log 2\pi + \frac{1}{12k} + O\left(\frac{1}{k^2}\right).$$
We obtain $(2)$ by ignoring all terms of order less than $n^2$ in Stirling's approximation. For $(3)$, a comparison with the integral $\int_1^n t\log t\,dt$ yields
$$\sum_{k=1}^n k\log k = \frac{n^2}{2}\log n - \frac{n^2}{4} + O(n\log n),$$
and hence
$$\log \prod_{k=1}^n k! = \sum_{k=1}^n (k\log k - k) + O(n\log n) = \frac{n^2}{2}\log n - \frac{3n^2}{4} + O(n\log n).$$
