Let $1\lt k \in\mathbb{N}. $ Prove: $\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k}\lt ln(k) \lt 1+ \frac{1}{2}+...+\frac{1}{k-1}$ using integration.

Question

Let $1\lt k \in\mathbb{N}. $ Prove: $\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k}\lt ln(k) \lt 1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-1}$ using definite integrals.

Hey everyone. I've been trying to prove this statement by splitting the interval $[1,k]$ into $k$ intervals of 1 length, and approximating the definite integral $\int^k_1 (\frac{1}{x})dx $, but am having issues. I would be happy to get some help on this question. Thanks in advance :)

Answer 1

Hint: Try fit in some rectangles (see @Samuel's comment)