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

Question

I am not sure how to approach this question. I tried to move the limit into the sum, but that turns the result to $0$, which is incorrect.

Answer 1

Denote $H_n = \sum_{k=1}^n \frac{1}{k}$ is the n-th harmonic series. We will use the asymptotic form of $H_n$. Therefore: $$\lim_{n \to \infty} \sum_{k=n}^{2n} \frac{1}{k} = \lim_{n \to \infty} H_{2n} - H_n =\lim_{n \to \infty} \ln (2n) + \gamma~ - (\ln n + \gamma) = \ln 2 $$