Answer
Follows from $\color{blue}{R_{n-1}} < \ln n < \color{green}{L_{n-1}}$, where $R_{n-1}$ and $L_{n-1}$ are, respectively, the right-endpoint and left-endpoint Riemann sums taken on a partition of $[1,n]$ into $n-1$ congruent subintervals, these Riemann sums approximating the integral $\displaystyle \int_1^n 1/x\ dx = \ln n$.
The diagram shows $\color{blue}{\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4}} < \ln 4 < \color{green}{1 + \dfrac{1}{2} + \dfrac{1}{3}}$. See details below for the proof for any integer $n\ge 2$.
Work Step by Step
We show that for integer $n\ge 2$, the left-hand and right-hand side of the given compound inequality are, respectively, the right-endpoint ($R_{n-1}$ ) and left-endpoint ($L_{n-1}$) Riemann sums that approximate $\displaystyle \ln n = \int_1^n 1/x\ dx$, this definite integral being the area under the curve $1/x, x\gt 0$, over the closed interval $[1,n]$. Since $f(x)=1/x,x\gt 0,$ is a monotonically decreasing function, the left-endpoint Riemann sum $L_{n-1}$ overestimates $\ln n$ while the right-endpoint Riemann sum underestimates $\ln n$, so that the $R_{n-1} < \displaystyle \ln n = \int_1^n 1/x\ dx < L_{n-1}$. In fact, each approximating rectangle of $L_{n-1}$ overestimates the corresponding subarea under the curve $y=1/x, 1\le x\le n,$ that it is estimating. Similarly, each approximating rectangle of $R_{n-1}$ underestimates the corresponding subarea under the curve $y=1/x, 1\le x\le n,$ that it is estimating.
(See diagram for an illustration for the case $n=4$, $\color{blue}{R_3} < \ln 4 < \color{green}{L_3}$ over the interval $[1,4]$).
$\underline{L_{n-1}}\ \text{(left endpoints)}$
function: $f(x) = 1/x, x\gt 0$
closed interval: $[a,b]= [1,n]$
number of congruent subintervals: $n-1$
width of subinterval: $\Delta x = \dfrac{b-a}{n-1} = \dfrac{n-1}{n-1} = 1$
partition: $1 < 2 < 3 < \ldots < n-2 < n-1$
$\begin{align*}
L_{n-1} &= \sum_{k=0}^{n-1} f\left(a + k\cdot \Delta x \right) \cdot \Delta x \\
&= \sum_{k=0}^{n-1} \frac{1}{1 +k\cdot (1)} \cdot 1 \\
&= \frac{1}{1+(0)(1)} + \frac{1}{1+(1)(1)} + \frac{1}{1+(2)(1)} + \ldots + \frac{1}{1+(n-1)(1)} \\
\color{green}{L_{n-1}}\ &\color{green}{= 1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{n-1}.}
\end{align*}$
$\underline{R_{n-1}}\ \text{(right endpoints)}$
function: $f(x) = 1/x, x\gt 0$
closed interval: $[a,b]= [1,n]$
number of congruent subintervals: $n-1$
width of subinterval: $\Delta x = \dfrac{b-a}{n-1} = \dfrac{n-1}{n-1} = 1$
partition: $2 < 3 < 4 < \ldots < n-1 < n$
$\begin{align*}
R_{n-1} &= \sum_{k=1}^n f\left(a + k\cdot \Delta x \right) \cdot \Delta x \\
&= \sum_{k=1}^n \frac{1}{1 +k\cdot (1)} \cdot 1 \\
&= \frac{1}{1+(1)(1)} + \frac{1}{1+(2)(1)} + \frac{1}{1+(3)(1)} + \ldots + \frac{1}{1+(n)(1)} \\
\color{blue}{R_{n-1}}\ &\color{blue}{= \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{n}.}
\end{align*}$
Now, since $\color{blue}{R_{n-1}} < \displaystyle \int_1^n 1/x\ dx < \color{green}{L_{n-1}}$, as we have discussed above (see diagram for an illustration of the case $n=4$), it follows that
$\color{blue}{\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \ldots + \dfrac{1}{n}} < \ln n < \color{green}{1+\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \ldots + \dfrac{1}{n-1}}$.
