Function approximating this product 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.
 A: 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

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

A: 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
