Limit of $x_n=\sum_{k=np+1}^{nq}\frac{1}{k}$ using Riemann sum

Question

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!

Answer 1

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.

Answer 2

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

Answer 3

Some additional transformation have indeed to be made: you got $x_n=\frac{1}{n}\sum_{k=1}^{n(q-p)}\frac{1}{p+\frac{k}{n}}$; we can continue by writing $$ x_n=\frac{q-p}{n\left(q-p\right)}\sum_{k=1}^{n(q-p)}\frac{1}{p+\frac{k\left(q-p\right)}{n\left(q-p\right)}}= \left(q-p\right)\frac 1{N_n}\sum_{k=1}^{N_n}f\left(\frac{k}{N_n}\right) $$ where $N_n=n (q-p)$ and $f\colon x\mapsto 1/\left(p+x(q-p)\right)$.