Function approximating this product

Question

Is there any function approximating, for large values of $p$, the quotient between the product of all primes and the product of all primes $-1$?

Basically: $2/1 \cdot 3/2 \cdot 5/4 \cdot 7/6 \cdot 11/10 \cdots $

Can that be approximated, for large values of $p$, with some known function?

Thank you very much.

Answer 1

To find the behavior of

$$ \prod_{p \leq x} \frac{p}{p-1} = \prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} $$

We begin by taking its logarithm, which we then rewrite as

$$ \log \prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} = - \sum_{p \leq x} \log\left(1 - \frac{1}{p}\right). $$

Now $\log(1-1/p) \sim -1/p$ for large $p$, so this should yield a good first approximation. After pulling this out we obtain a sum which converges, so let's rewrite the quantity as

$$ \begin{align} &- \sum_{p \leq x} \log\left(1 - \frac{1}{p}\right) \\ &\qquad = \sum_{p \leq x} \frac{1}{p} - \sum_{p \leq x} \left[\log\left(1 - \frac{1}{p}\right) + \frac{1}{p}\right] \\ &\qquad = \sum_{p \leq x} \frac{1}{p} - \sum_{p} \left[\log\left(1 - \frac{1}{p}\right) + \frac{1}{p}\right] + \sum_{p > x} \left[\log\left(1 - \frac{1}{p}\right) + \frac{1}{p}\right]. \tag{$*$} \end{align} $$

According to Mertens' formula (see the second entry here, where $M$ is the Meissel-Mertens constant),

$$ \sum_{p \leq x} \frac{1}{p} - \sum_{p} \left[\log\left(1 - \frac{1}{p}\right) + \frac{1}{p}\right] = \log\log x + \gamma + o\left(\frac{1}{\log x}\right). $$

Using Mertens' formula again with the Abel summation formula it's possible to show that the final sum in $(*)$ is also $o\left(\frac{1}{\log x}\right)$, allowing us to conclude that

$$ -\sum_{p \leq x} \log\left(1 - \frac{1}{p}\right) = \log\log x + \gamma + o\left(\frac{1}{\log x}\right). $$

Exponentiating this we find that

$$ \prod_{p \leq x} \left(1 - \frac{1}{p}\right)^{-1} = e^\gamma \log x + o(1). $$

If you'd like, you can replace $x$ by $p_n$, the $n^\text{th}$ prime, and use the estimate

$$ p_n \approx n\log n + n\log\log n - n + \cdots $$

to see that the product of the first $n$ primes is

$$ \begin{align} \prod_{k=1}^{n} \left(1-\frac{1}{p_k}\right)^{-1} &= e^\gamma \log p_n + o(1) \\ &\approx e^\gamma \log\Bigl(n \log n + n\log\log n - n\Bigr). \end{align} $$

Answer 2

You are asking for an approximation of $$\prod_{p\leq n} \frac{p}{p-1}$$
It is well known that $\prod_{p\leq n} \frac{p}{p-1} \simeq lnn$.
For more details and to learn how Euler used this fact to prove something really interesting see here