Show that ${A_n -1 \le} \ln(n) \le A_{n-1}$ for any positive integer $n$, where $${A_n=} \sum_{i=1}^n \frac{1}{i} $$
I tried splitting the interval into small rectangles with the upper sum and lower sum but I still don't understand how can you achieve this ${A_n -1 \le} \ln(n) \le A_{n-1}$
On
We know that $$\int _1 ^n \frac 1 x dx \lt A_{n-1} $$ Intuitively, the right side is a series of rectangle with width $1$ and length from $1$ to $\frac 1 {n-1}$
However it is also true that $$\int _1 ^n \frac 1 x dx \gt A_n -1 $$. This time, the right side is a series of rectangle with width $1$ but length from $\frac 1 2$ to $\frac 1 n$.
Therefore $${A_n -1 \le} \ln(n) \le A_{n-1}$$
Let $$I(n) = \int_1^n \frac{1}{x} \; dx = \log n$$ Now split the interval $[1, n]$ into $n-1$ unit length intervals $[1,2]$, $[2, 3]$, etc, to give $$I(n) = \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{x} \; dx$$ Note that $\frac{1}{x}$ is decreasing, so for $x \in [k, k+1]$, we have $$\frac{1}{k+1} \leq \frac{1}{x} \leq \frac{1}{k}$$ We use this to bound each integral, so that $$I(n) \leq \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{k} \; dx = \sum_{k=1}^{n-1} \frac{1}{k} = A_{n-1}$$ and similarly $$I(n) \geq \sum_{k=1}^{n-1} \int_k^{k+1} \frac{1}{k+1} \; dx = \sum_{k=2}^{n} \frac{1}{k} = A_n - 1$$ Putting these together, we have $$A_n - 1 \leq I(n) \leq A_{n-1}$$ which gives the result we need, since $I(n) = \log n$.