Show that $\lim_{n\to\infty}\sum_{n=k}^{2n}\frac{1}{k} = \ln 2$

Question

I would like to show that $$\lim_{n\to\infty}\sum_{k=n}^{2n}\frac{1}{k} = \ln 2.$$ Any tips on how to go about it? I feel like I should use the squeeze theorem but I fail to find convergent lower and upper bounds.

Answer 1

hint

$$\int_k^{k+1}\frac{dt}{t}\le \frac 1k\le \int_{k-1}^k\frac{dt}{t}$$

$$\int_{n}^{2n+1}\frac{dt}{t}\le \sum_{k=n}^{2n}\frac 1k\le \int_{n-1}^{2n}\frac{dt}{t}$$