Convert Riemann Sum into definite integral

Question

I have the equation $$\lim_{n\to\infty}\frac{\left(r-1\right)}{n}\sum_{j=r}^{n}\frac{n}{j-1}\left(\frac{1}{n}\right)$$ which is converted to a definite integral using x as the limit of (r-1)/n, t for (i-1)/n and dt for 1/n

$${x\int _{x}^{1}{\frac {1}{t}}\,dt}.$$

I can only make sense of this intuitively but cannot convert it myself. Can someone show me the step by step process of how they converted the equation using the appropriate substitutions of a and b in the following equation.

$${\int _{x=a}^{b}{f(x)}\,dx}=\lim_{n\to\infty}\sum_{j=1}^{n}f\left(a+\frac{b-a}{n}j\right)\frac{b-a}{n}.$$

One of the main things confusing me is that j is not equal to 1 or 0 unlike typical Riemann sums. Any help will be greatly appreciated!

Answer 1

The sum written is $$\frac{\left(r-1\right)}{n}\sum_{j=r}^{n}\frac{n}{j-1}\left(\frac{1}{n}\right) = (r-1)\sum_{j=r}^{n}\dfrac{1}{n(j-1)} = \dfrac{r-1}{n}\cdot\Big(H_{n-1}-H_{r-1} \Big)$$ And this limit evaluates to $0$ as $\lim_{n \to \infty}\dfrac{\ln(n)}{n} = 0$

Answer 2

$$L=\lim_{n\to\infty}\frac{\left(r-1\right)}{n}\sum_{j=r}^{n}\frac{n}{j-1}\left(\frac{1}{n}\right)$$ $$L=\lim_{n \to \infty}\frac{(r-1)}{n}\sum_{j=r}^{n}\frac{n}{j-1}\left(\frac{1}{n}\right)=\lim_{n \to \infty}\frac{(r-1)}{n}\int_{r/n}^{1}\frac{dx}{x}=\lim_{n \to \infty}\frac{(r-1)}{n}\ln(n/r)=0.$$ Lastly L-Hospiltal rule has been used.