Is there a close approximation for $\prod_{i=1}^n (1-1/p_i)$?

Question

Consider the primes $p_i$ starting at $p_1 = 2$. Is there a close approximation for

$$\prod_{i=1}^n (1-1/p_i)\;?$$

Plotting the values you get the following.

Answer 1

You've discovered Merten's 3rd theorem (https://en.wikipedia.org/wiki/Mertens%27_theorems)!

This states, in your notation, that

$$\prod_{i=1}^{n}\left(1-\frac{1}{p_i}\right)\sim \frac{e^{-\gamma}}{\ln(p_n)}.$$

Here, $"\sim"$ means that the two ratio between these two quantities approaches $1$ as $n$ goes to infinity, $e$ is Euler's constant, and $\gamma$ is the Euler-Mascheroni constant. There are stronger error bounds, assuming stronger versions of the Prime Number Theorem.

If you don't know the Prime Number Theorem, it states that $p_n\sim n\ln(n)$. By "stronger versions", I mean some version of this theorem where we bound the error between $p_n$ and $n\ln(n)$. Using this theorem, we can restate Merten's theorem as

$$\prod_{i=1}^{n}\left(1-\frac{1}{p_i}\right)\sim \frac{e^{-\gamma}}{\ln(n\ln(n))}=\frac{e^{-\gamma}}{\ln(n)+\ln(\ln(n))}.$$

However, since $\ln(\ln(n))$ is small relative to $\ln(n)$ this can equivilantly be seen as

$$\prod_{i=1}^{n}\left(1-\frac{1}{p_i}\right)\sim \frac{e^{-\gamma}}{\ln(n)}.$$

Answer 2

Let us consider \begin{align*} \prod_{i=1}^\infty \left(1-\dfrac{1}{p_i}\right)&=\left(\prod_{i=1}^\infty \dfrac{p_i }{p_i-1}\right)^{-1}=\left(\prod_{i=1}^\infty \dfrac{1}{1-p_i^{-1}}\right)^{-1}=\zeta(1)^{-1}=\left(\sum_{n=1}^\infty\dfrac{1}{n} \right)^{-1} \end{align*} This series diverges, i.e. tends to infinity, so its inverse tends to zero. Your question can therefore be rephrased as to how fast the harmonic series increases. One possible answer to that is $$ \int_1^{n+1}\dfrac{dx}{x}<\sum_{i=1}^n \dfrac{1}{i}<1+\int_1^{n}\dfrac{dx}{x} $$