Limit of $\frac1x \sum_{z=1}^{z=x}\ln(z)$ as $x$ approaches infinity As I was reading a book on market microstructure, I came across an equation:
$\text{Price} = m_0 + \frac{2G_1(1+\gamma)}{Q}\sum_{L=1}^{Q}\ln L$. 
The book then proceeds to say, for $Q\gg 1$: 
$\text{Price }\approx m_0 + 2G_1(1+\gamma)(\ln Q - 1)$
Hence, the book implies $\frac{\sum_{z=1}^{x}\ln (z)}{x} \to \ln(x)-1$ as $x \to \infty$.
So far, I assumed that as $\sum_{z=1}^{x}\ln(z) \to \ln(x)$ as  $x \to \infty$.
Assuming that is correct, I am having trouble for the final step. How does $\frac{\ln(x)}{x} \to\ln x-1$ ? And any help will go a long way. Thank you for your time and help!
 A: Using Abel's summation formula we have $$\sum_{n\leq x}\log\left(n\right)=\sum_{n\leq x}1\cdot\log\left(n\right)=x\log\left(x\right)-\int_{1}^{x}\frac{\left\lfloor t\right\rfloor }{t}dt$$ where $\left\lfloor x\right\rfloor $ is the floor function and so, using $\left\lfloor x\right\rfloor =x+O\left(1\right)$, we get $$\sum_{n\leq x}\log\left(n\right)=x\log\left(x\right)-x+O\left(\log\left(x\right)\right)$$ hence $$\frac{1}{x}\sum_{n\leq x}\log\left(n\right)=\color{red}{\log\left(x\right)-1+o\left(1\right)}.$$
A: Note that, since $\ln(x) + \ln(y) = \ln(xy)$, you have
$$\begin{equation}\begin{aligned}
\sum_{z=1}^{x} \ln (z) & = \ln(1) + \ln(2) + ... + \ln(x) \\
& = \ln((1)(2)(3)\cdots (x-1)(x)) \\
& = \ln(x!)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Stirling's approximation gives that
$$\ln(x!) = x\ln(x) - x + O(\ln(x)) \tag{2}\label{eq2A}$$
You thus have, since $\lim_{x \to \infty}\frac{\ln(x)}{x} = 0$, that
$$\frac{\sum_{z=1}^{x} \ln (z)}{x} = \frac{x\ln(x) - x + O(\ln(x))}{x} \to \ln(x) - 1 \; \text{ as } \; x \to \infty \tag{3}\label{eq3A}$$
A: In many cases, the sums $\sum_{k=1}^n f(k)$ can be approximated with $\int_1^n f(x) dx $. 
Another approach:
$$\frac{1}{n}\sum_{k=1}^n \log k -  \log n= \frac{1}{n}\sum_{k=1}^n \log \frac{k}{n}$$
and the latter is a Riemann sum for the function $\log x$ on $(0,1]$, so the limit will be the (improper) Riemann integral of this function $\int_{0_{+}}^1 \log x\, dx = -1$.
