Estimate for a prime product.

Question

Is there a bound for $$\prod_{i=1}^{m}\Big(1-\frac{1}{p_i}\Big)$$ where $p_i$ is $i$th prime?

What if $m=O(\log n)$?

Answer 1

Let $n_i$ be such that $p_i^{n_i}\ge p_{m+1}$ For all $n$ we have $\frac1{1-\frac1{p_i}}\ge 1+p_i^{-1}+\ldots+p_i^{-n}$. As every natural number, less than $p_{m+1}$ can be uniquely written as product of powers of $p_i$s, $$\prod_{i=1}^m \left(\frac1{1-\frac1{p_i}}\right)\ge \sum_{i=1}^{p_{m+1}-1} \frac1i\ge 1+\log p_{m+1}$$ So, your product is can be bounded by $$\prod_{i=1}^m \left(1-\frac1{p_i}\right)\le \frac1{1+\log p_{m+1}}$$ Using PNT, $\log p_{m+1}\sim \log m-\log\log m$. So, $m\sim \log n$ gives $\log p_{m+1}\sim \log\log n-\log\log\log n$

Answer 2

Hint

It is known that $$\dfrac{\phi(n)}n = \prod\limits_{p\,|\, n}\left(1-\dfrac1p\right).$$ Required production equals to $\dfrac{\phi(p_n\#)}{p_n\#}.$
And I can recommend light and detail info.