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




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.


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 $$\sum_{n<p\leq p_{n}}\log\left(\frac{p}{p-1}\right)\sim 0.$$

  • $\begingroup$ 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 $\endgroup$ – Will Jagy Sep 12 '14 at 18:58

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