Limit of a Riemann Sum.

Question

I am trying to calculate the limit

$$\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac{1}{k}$$

Can someone please explain how I can go about doing this?

Answer 1

Let $S_n=\sum\limits_{n+1}^{2n}\frac{1}{k}$

$\int_{k-1}^k\frac{dx}{x}\gt \frac{1}{k}\gt \int_k^{k+1}\frac{dx}{x}$

$\int_n^{2n}\frac{dx}{x}\gt S_n \gt \int_{n+1}^{2n+1}\frac{dx}{x}$

$ln(2)\gt S_n\gt ln(2-\frac{1}{n+1})$ squeeze $\lim{n\to \infty}\ S_n=ln(2)$.