Understanding how to form a definite integral from limit of an infinite sum I have to solve the limit $$\lim_{n\rightarrow\infty}\left(\frac1n\sum_{k=1}^n(\text{log}\:k)^2-\left(\frac1n\sum_{k=1}^n(\text{log}\:k)\right)^2\right)$$
I don't know how to proceed in this problem. However, the solution says that this limit is equal to the limit $$\lim_{n\rightarrow\infty}\left(\frac1n\sum_{k=1}^{n-1}\left(\text{log}\:\frac kn\right)^2-\left(\frac1n\sum_{k=1}^{n-1}\left(\text{log}\:\frac kn\right)\right)^2\right)$$
Which is equal to $$\int_0^1 (\text{log}\:x)^2dx-\left(\int_0^1 \text{log}\:x\;dx\right)^2$$
Can someone explain how these two equations were derived from the first expression? I am aware of expressing definite integral as limit of an infinite sum. But I don't see how the above conversions were made. I need the intermediate steps.
 A: Write
$$\log k = \log k - \log n + \log n = \log \frac{k}{n} + \log n,$$
so that 
\begin{align}\left(\frac{1}{n}\sum_{k = 1}^n \log k\right)^2 &= \left(\frac{1}{n}\sum_{k = 1}^n \log\frac{k}{n} + \log n\right)^2\\
&= \left(\frac{1}{n}\sum_{k = 1}^n \log \frac{k}{n}\right)^2 + \frac{2\log n}{n}\sum_{k = 1}^n \log \frac{k}{n} + (\log n)^2 \tag{1}
\end{align}
and
\begin{align}\frac{1}{n}\sum_{k = 1}^n (\log k)^2 &= \frac{1}{n}\sum_{k = 1}^n \left[\left(\log\frac{k}{n}\right)^2 + 2\log n\log\frac{k}{n} + (\log n)^2\right]\\
&=\frac{1}{n}\sum_{k = 1}^n \left(\log \frac{k}{n}\right)^2 + 2\frac{\log n}{n}\sum_{k = 1}^n \log \frac{k}{n} + (\log n)^2 \tag{2}
\end{align}
Subtracting $(1)$ from $(2)$, we obtain
\begin{align}
& \frac{1}{n}\sum_{k = 1}^n (\log k)^2 - \left(\frac{1}{n}\sum_{k = 1}^n \log k\right)^2 \\
&= \frac{1}{n}\sum_{k = 1}^n \left(\log \frac{k}{n}\right)^2 - \left(\frac{1}{n}\sum_{k = 1}^n \log \frac{k}{n}\right)^2\\
&= \frac{1}{n}\sum_{k = 1}^{n-1} \left(\log \frac{k}{n}\right)^2 - \left(\frac{1}{n}\sum_{k = 1}^{n-1} \log \frac{k}{n}\right)^2
\end{align}
The last step follows since $\log \frac{k}{n} = 0$ when $k = n$.
Since 
$$\frac{1}{n}\sum_{k = 1}^{n-1} \log \frac{k}{n}$$
is a sequence of Riemann sums for $\log x$ over $[0,1]$, and 
$$\frac{1}{n}\sum_{k = 1}^{n-1} \left(\log \frac{k}{n}\right)^2$$
is a sequence of Riemann sums for $(\log x)^2$ over $[0,1]$, we have
$$\frac{1}{n}\sum_{k = 1}^{n-1} \log \frac{k}{n} \to \int_0^1 \log x\, dx$$
and 
$$\frac{1}{n}\sum_{k = 1}^{n-1} \left(\log \frac{k}{n}\right)^2 \to \int_0^1 (\log x)^2\, dx,$$
whence the third step follows.
