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.
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.
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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.