How to prove that for $n \in \mathbb{N}$ we have $\sum_{k=2}^n \frac{1}{k}\leq \ln(n) \leq \sum_{k=1}^{n-1} \frac{1}{k}$

Question

Prove that for $n \in \mathbb{N}$ we have $\sum_{k=2}^n \frac{1}{k}\leq \ln(n) \leq \sum_{k=1}^{n-1} \frac{1}{k}$ by using Riemann integral?

Answer 1

Observe that $\frac{1}{n}$ is decreasing $(1)$.

$$\int\limits_{1}^{n}{\dfrac{1}{x} \, dx}=\int\limits_{1}^{2}{\dfrac{1}{x}\, dx}+\int\limits_{2}^{3}{\dfrac{1}{x}\, dx}+\cdots +\int\limits_{n-1}^{n}{\dfrac{1}{x}\, dx} \tag{2}$$

For an integrable function $f$, if $ m \le f(x)\le M \quad \forall x\in [a,b]$ then $$m(b-a)\le\int\limits_{a}^{b}{f(x) dx} \le M(b-a) \tag 3$$

Observe with the help of $(1)$ and $(3)$ that $$\dfrac{1}{m}\le\int\limits_{m-1}^{m}{\dfrac{1}{x}\, dx}\le \dfrac{1}{m-1} \tag{4}$$

Using $(2)$ and $(4)$ see that $\sum_{k=2}^n \frac{1}{k}\leq \ln(n) \leq \sum_{k=1}^{n-1} \frac{1}{k}$

Answer 2

Hint. Note that on $[k, k+1] $ we have $$ \frac 1{k+1} \le \frac 1x \le \frac 1k $$ Now integrate over $[k, k+1] $ and sum from 1 to $ n-1$.