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!

We just have to calculate $\int_0^{q-p}$ instead of $\int_0^1$. Then, with
$$f(x)=\frac{1}{p+x}$$
$$\int_0^{q-p}\frac{1}{p+x}dx=[ln|p+x|]_0^{q-p}=ln|p+q-p|-ln|p|=ln\frac{q}{p}$$
And we are done.
