A question about the convergence of partial products of zeta of one.

Recently I've been toying around with the Totient function and the Prime Number Theorem and came up with the odd result that the following limit

$$\lim_{n\to\infty}\frac{\pi(n)m_n}{\phi(m_n)n}$$

where

$$m_n\equiv\prod_{i=1}^{n}p_i$$

where $p_i$ is the $i$th prime converges to approximately $e$ (somewhat verified computationally).

This is, by the prime number theorem, equivalent to the statement that

$$\lim_{n\to\infty}\prod_{i=1}^{n}\left(\frac{p_i}{p_i-1}\right)/\log n\approx e$$.

Or, taking the logarithm,

$$\lim_{n\to\infty}\left[\sum_{i=1}^{n}\log\left(\frac{p_i}{p_i-1}\right)-\log\log n\right]\approx 1$$

Furthermore, by a relation involving the Euler-Mascheroni constant, this implies

$$\lim_{n\to\infty}\sum_{n<p\leq p_n}\log\left(\frac{p}{p-1}\right)\approx 1-\gamma$$

My question is whether or not anyone knows how to prove any of these statements, either with equality (perhaps computation is leading me astray) or with a different value as the limit.

1 Answer

Some of your formula's are not quite correct. It is a result of Mertens that $$\prod_{p\leq x} \left(1-\frac{1}{p}\right)^{-1} \sim e^\gamma \log x,$$ (see Merten's Third Theorem on this page) and we are taking $x=p_n \sim n\log n$ so it follows that

$$\lim_{n\rightarrow \infty} \frac{n}{\phi(n)} \frac{\pi(n)}{n}=e^\gamma.$$ Consequently

$$\lim_{n\rightarrow \infty} \left[ \sum_{i=1}^n \log\left(\frac{p_i}{p_i-1}\right)-\log \log n\right]=\gamma.$$

Now, using the fact that $$\sum_{p\leq x}\log\left(\frac{p}{p-1}\right)\sim\log\log x+\gamma$$ and $p_n\sim n\log n$ we find that $$\sum_{n<p\leq p_{n}}\log\left(\frac{p}{p-1}\right)\sim 0.$$

• Dropped answer, redundant. Jean-Louis Nicolas 1983, RH is equivalent to : $$\frac{P}{\phi(P)} > e^\gamma \log \log P$$ for all primorials $P.$ Journal of Number Theory, vol. 17, pages 375-388 – Will Jagy Sep 12 '14 at 18:58