I'm trying to understand the proof outlined in this question. There are related questions, but those concern different parts of the proof.
For completeness, here it is:
I have trouble understanding the inequality
$$1 + \frac{1}{2} + \ldots + \frac{1}{n} \leq \sum \frac{1}{m}$$
with the sum spanning over all $m\in\mathbb{N}$ which only contain prime factors that are less than $x$.
How can this be shown?
All of the numbers $1,2,\ldots,n$ are $\leq x$ (since $n \leq x$), and thus have only prime divisors $\leq x$ (since a divisor of a positive integer $m$ is always $\leq m$). Thus, each addend of the sum $\dfrac{1}{1} + \dfrac{1}{2} + \cdots + \dfrac{1}{n}$ is also an addend on the sum $\sum \dfrac{1}{m}$ which extends over all positive integers $m$ which have only prime divisors $\leq x$. The latter sum may have additional addends, but they are nonnegative, so they can only make the sum bigger. Hence, $\dfrac{1}{1} + \dfrac{1}{2} + \cdots + \dfrac{1}{n} \leq \sum \dfrac{1}{m}$.