Limit of $x_n=\sum_{k=np+1}^{nq}\frac{1}{k}$ using Riemann sum I am trying to find the limit of the following sequence using Riemann sum:
$$x_n=\sum_{k=np+1}^{nq}\frac{1}{k}\qquad p,q\in\mathbb{N}\quad p<q$$
I have tried to develope the expression:
$$\frac{1}{np+1}+\frac{1}{np+2}+...+\frac{1}{nq}=\frac{1}{n}(\frac{1}{p+\frac{1}{n}}+\frac{1}{p+\frac{2}{n}}+...+\frac{1}{p+\frac{n(q-p)}{n}})=\frac{1}{n}\sum_{k=1}^{n(q-p)}\frac{1}{p+\frac{k}{n}}$$
But I need an expression like $\frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})$, from $k=1$ to $n$, not to $n(q-p)$, so I can calculate the limit as $\int_0^1f(x)dx.$
Could you give me some hints? Thanks!

$$\lim_{n\to \infty}x_{n}{=\lim_{n\to \infty}\sum_{k=np}^{nq+1}{1\over k}\\=\lim_{n\to \infty}\sum_{k=np}^{nq+1}{n\over k}\cdot {1\over n}\\=\lim_{n\to \infty}\sum_{k=np}^{nq+1}{1\over{k\over n}}\cdot {1\over n}\\=\int_p^q{dx\over x}\\=\ln {q\over p}}$$
