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Let p is prime. In this question and answer (Intuition for the prime number theorem）, they approximate $ \prod_{p\leq n} (1-1/p)^{-1} \approx \sum_{i=1}^n 1/i$. This is well known way to get the intuition of prime number theorem. However, I'm not convinced well.

How to know that the order of $$ \left( \prod_{p\leq n} (1-1/p)^{-1} - \sum_{i=1}^n 1/i \right)$$ is sufficiently small? In other words, $O(\log n)$.

ueir
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  • Intuition means "the power or faculty of attaining to direct knowledge or cognition without evident rational thought and inference" (https://www.merriam-webster.com/dictionary/intuition), so there can't be a "method" for it by definition. $\Theta(\log n)$ or $o(\log n)$ aren't limits, but orders of magnitude. Suggestion: you might consider learning and applying definitions instead of "intuition". –  Oct 09 '20 at 11:13
  • I change the statement a little according to you. – ueir Oct 09 '20 at 11:24
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    I wouldn't call that "small". This is the same magnitude as the quantities being subtracted. – metamorphy Oct 09 '20 at 11:59
  • I mean, sufficiently small so that prime number thoerem would hold. – ueir Oct 09 '20 at 12:17

1 Answers1

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Mertens' estimate gives

$$ \prod_{p\leq n}(1-1/p)^{-1} = C\log n + O(1) $$

where $C=e^{\gamma} = 1.78\cdots$. Comparing this to the elementary estimate

$$ \sum_{i\leq n}\frac{1}{i} = \log n + O(1),$$

we see that in fact the difference

$$ \prod_{p\leq n} (1-1/p)^{-1}-\sum_{i\leq n}\frac{1}{i} = (C-1)\log n+O(1),$$

so can give a precise asymptotic formula for the size of the difference. In particular, the difference is $O(\log n)$ (and this is the correct order of magnitude).

metamorphy
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Thomas Bloom
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