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.