Let$\ p_n$ be the$\ n$-th prime. Is$\ \lim_{n\to\infty} \log \log n \prod_{i=1}^{\lfloor \log n \rfloor} \frac{p_i-1}{p_i}>0$?

Question

I'm less than a novice in analysis, I don't even know how to approach this. Thanks in advance.

Answer 1

The limit is $e^{-\gamma}$ where $\gamma$ is the Euler–Mascheroni constant. This is a direct consequence of

Mertens' $3^{rd}$ theorem $$\lim_{x\to\infty} \log x \prod_{\substack{p\,\le\,x\\p \text{ prime}} }\left(1-\frac{1}{p}\right) = e^{-\gamma}$$

By Prime number theorem, we know $p_n \sim n\log n$, so $$\lim_{n\to\infty}\frac{\log n}{\log p_n} = 1$$

If we substitute $x$ by $p_n$ in Mertens' theorem and multiply its LHS by $\displaystyle\;\frac{\log n}{\log p_n}$, we get

$$ \lim_{n\to\infty} \log n \prod_{i=1}^n \left(1 - \frac{1}{p_i}\right) = \lim_{n\to\infty} \frac{\log n}{\log p_n} \left[ \log p_n \prod_{i=1}^n \left(1 - \frac{1}{p_i}\right)\right]\\ = \lim_{n\to\infty} \log p_n \prod_{\substack{p\,\le\,p_n\\p\text{ prime}}}\left(1 - \frac{1}{p}\right) = e^{-\gamma}$$

For a proof of Mertens' theorem, see

• G.H Hardy's and E.M. Wright,
  Chapter XXII - The Series of Primes (3),
  An Introduction to the Theory of Numbers

It is theorem $429$ there.

Answer 2

No, because as $n \to \infty$, we have that $x_n=\prod_{i=1}^{\log n} \frac{p_i-1}{p_i}=\prod_{i=1}^{k} \frac{p_i-1}{p_i}$ diverges*, where $k=\log n \to \infty$. Hence, we have:

$$\displaystyle \lim_{n \to \infty}\log \log nx_n=\log \log \infty=\infty$$

*(by your other question Let$\ p_n$ be the$\ n$-th prime. Can you give me a proof for$\ \prod_{i=1}^\infty \frac{p_i-1}{p_i}=P\approx \frac{1}{11.0453}$?)

Answer 3

First note that, for every $u$ in $(0,\frac12]$, $$-u^2\leqslant\log(1-u)+u\leqslant0,$$ and the series $$\sum\limits_p\frac1{p^2}\leqslant\sum\limits_n\frac1{n^2}$$ converges hence, when $x\to\infty$, $$\sum\limits_{p\lt x}\log\left(1-\frac1p\right)=c+o(1)-\sum\limits_{p\lt x}\frac1p,$$ for some constant $c$ On the other hand, the difference $$\sum\limits_{p\lt x}\frac1p-\log\log x,$$ converges to the so-called Meissel-Mertens constant $M$, hence, considering the exponentials, $$ \prod\limits_{p\lt x}\left(1-\frac1p\right)=\mathrm e^{c+o(1)}\,\mathrm e^{-M}\,\frac1{\log x},$$ in particular, $$\lim_{x\to\infty}\log x\cdot\prod\limits_{p\lt x}\left(1-\frac1p\right)=\mathrm e^{c-M}\gt0.$$ The same result holds for the limit along the sequence $x=\lfloor\log n\rfloor$.