Can someone please tell me why the last claim "It is thus the case…" is true?
I tried considering negation of the last claim. But it didn't help.
Any help would be appreciated. Thanks in advance.
analytic-number-theorynumber theoryprime numbersprime-gaps
Best Answer
The product on the right consists of the sum of $\frac 1n$ where $n$ is any number divisible only by the primes $≤m$. These numbers may occur with some multiplicity, that doesn't matter. It is also easy to see that the product on the right is $$\frac 2{2-1}\times \frac 3{3-1}\times \cdots \times \frac m{m-1}=m$$
Thus, if every natural number from $1$ to $4^m$ factored completely using the primes $≤m$ we'd have that the left hand sum was $≤ m$ contrary to the stated premise.