Proof the equation: $\lim\limits_{n\to\infty}\frac{(n+1)\ln(n!)-2\ln(2!3!\cdots n!)}{n^2+n}=\frac12$ I need some help to proof the next equation: 
$$\lim_{n\to\infty}\frac{(n+1)\ln(n!)-2\ln(2!3!\cdots n!)}{n^2+n}=\frac{1}{2}$$

$$\lim_{n\to\infty}\frac{(n+1)(\ln(2*Pi*n)/2 + n\ln(n) - n) -2(ln((2Pi)^n*n!)/2+\sum_{k=1}^nk\ln(k)-n(n+1)/2)}{n(n+1)}=\lim_{n\to\infty}\frac{(n+1)(\ln(2*Pi*n)/2 + n\ln(n) - n) - ln((2Pi)^n*n!)-2*\sum_{k=1}^nk\ln(k)}{n(n+1)}+1==\lim_{n\to\infty}\frac{\frac{-n}{2}\ln(2*Pi*n) + 2*\sum_{k=1}^nk(\ln(n)-\ln(k))}{n(n+1)} = \frac{1}{2}$$
 A: It helps to consider the terms separately and ignore everything that can't play any role for the limit since its order of growth is less than $O(n^2)$.
If we look at $(n+1)\ln n!$, Stirling's formula yields
$$\begin{align}
(n+1)\ln n! &= (n+1)\left[\left(n+\frac{1}{2}\right)\ln n - n + O(1)\right]\\
&= \left(n^2 + \frac{3}{2}n + \frac{1}{2}\right)\ln n - n(n+1) + O(n)\\
&= n^2\ln n - n(n+1) + O(n\ln n).
\end{align}$$
Looking at $2\ln (2!3!\dotsb n!)$, we obtain (adding a dummy factor $1! = 1$)
$$\begin{align}
2\ln (2!3!\dotsb n!) &= 2\sum_{k=1}^n \left[\left(k+\frac{1}{2}\right)\ln k - k + O(1)\right]\\
&= 2\sum_{k=1}^n k\ln k - n(n+1) + O(n\ln n).
\end{align}$$
We see that the only relevant terms in the numerator - the two $n(n+1)$ cancel - are
$$n^2\ln n - 2\sum_{k=1}^n k\ln k.$$
It remains to determine the sum with sufficient accuracy. We can now remove the $k = 1$ term that is $0$, and obtain
$$\sum_{k=2}^n k\ln k = \int_1^n t\ln t\,dt + \sum_{k=2}^n \left(\int_{k-1}^k k\ln k -  t\ln t\,dt\right).$$
On the interval $[k-1,k]$, the difference $k\ln k - t\ln t$ is smaller than $1+\ln k$, so
$$\sum_{k=2}^n k\ln k = \int_1^n t\ln t\,dt + O(n\ln n).$$
The integral can be explicitly evaluated,
$$\int_1^n t\ln t\,dt = \frac{n^2}{2}\ln n - \frac{n^2}{4} + \frac{1}{4},$$
and that takes us to
$$\frac{(n+1)\ln n! - 2\ln (2!3!\dotsb n!)}{n^2+n} = \frac{n^2/2 + O(n\ln n)}{n^2+n} \to \frac{1}{2}.$$
