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I was reading this article about the convergence of the series $\sum_p \frac{1}{p\cdot log(p)}$. The second comment uses a weaker version of Merten's First Theorems, but a got lost on the formula described as in the title of this question.

For each prime $p⩽n$, there are $k(p,n) := \left\lfloor\frac{n}{p}\right\rfloor$ multiples of $p$ that are $⩽n$, and hence $\prod_{p⩽n} p^{k(p,n)} ∣ n!$

I understood the account of the multiples of $p$. But, I can't understand the productory dividing $n!$ I appreciate any help :D

Igor Souza
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    If there are $m$ multiples of $p$ in ${1,2,3,\dots,n}$, then their product $n!$ must contain at least $m$ times the factor $p$, so at least the factor $p^m$ – Exodd Jun 16 '22 at 19:16
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    @wasn'tme you're not helping – Exodd Jun 16 '22 at 19:17
  • Sorry, I can't. You proved yourself that $p^m|n!$, and your $m$ is exactly $k(p,n)$. That's all, there isn't more to it. – wasn't me Jun 16 '22 at 19:22
  • ok, so $p^{k(n,p)}$ divides $n!$. What about the others $p \leq n$ on the productory? – Igor Souza Jun 16 '22 at 19:32
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    The reasoning above works for every $p$ prime less than $n$, so it must hold also for their product – Exodd Jun 17 '22 at 08:08

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