Prove the inequality between integral and summation of multiplicative inverse I want to prove the following inequality:

$$
\ln(n) = \int\limits_1^n{ \frac{1}{x} dx } \geq \sum_{x = 1}^{n}{\frac{1}{x + 1}} = \sum_{x = 1}^{n}{\frac{1}{x}} - 1
$$

I ask this question as I'm trying to calculate the upper bound of a harmonic series:

$$
\sum_{x = 1}^{n}{\frac{1}{x}} \leq \ln(n) + 1
$$
Using the mL (minumum times Length) estimate
$$\int_a^b |f(x)|dx\geq (b-a)\min_{x\in[a,b]}|f(x)|$$
One has
$$\int_1^n\frac{1}{x}dx=\sum_{k=1}^{n-1}\int_{k}^{k+1}\frac{1}{x}dx\geq\sum_{k=1}^{n-1}(k+1-k)\frac{1}{k+1}=\sum_{k=1}^{n-1}\frac{1}{k+1}=\sum_{k=1}^n\frac{1}{k}-1,$$
where I have used that $\frac{1}{x}$ is positive on the positive axis (so the absolute value is irrelevant) and that $\frac{1}{x}$ is decreasing (again, for positive $x$), so that
$$\min_{x\in[a,b]}\frac{1}{x}=\frac{1}{b}$$

PS: my suggestion to you, is to draw the situation: first draw the $\frac{1}{x}$ function, then subdivide the real axis in intervals of length $1$, and then build up the rectangles which fit under the curve, for each of the subintervals (bases). The sum of their areas will be given by the above formula.
The complete solution is provided here.