Prove that for $n\in \mathbb{N}$ we have $$\sum_{k=2}^{n} \frac{1}{k}\leq \ln(n)\leq \sum_{k=1}^{n-1} \frac{1}{k}$$
Here is what I know; $$\ln(n) = \int_1^n \frac{1}{k}dk$$ From examining the graph of function $1/n$, I can see the first inequality, as all the rectangles (length = 1, height = $1/n$) lie below the curve $1/n$. But i think I should show this more thoroughly. Also I can't figure how to think about second inequality. Appreciate your help.
$$ \ln n= \int_1^n\frac{1}{x}dx =\sum_{k=1}^{n-1}\int_{k}^{k+1}\frac{1}{x} dx$$ Now if $k\le x\le k+1$, then $\frac{1}{k+1}\le\frac{1}{x}\le \frac{1}{k}$, so $$\sum_{k=2}^n\frac{1}{k}=\sum_{k=1}^{n-1}\frac{1}{k+1} =\sum_{k=1}^{n-1}\int_k^{k+1}\frac{1}{k+1}dx \le \underbrace{\sum_{k=1}^{n-1}\int_{k}^{k+1}\frac{1}{x} dx}_{=\ln n} \le \sum_{k=1}^{n-1}\int_{k}^{k+1}\frac{1}{k} dx = \sum_{k=1}^{n-1}\frac{1}{k}$$