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For fixed $c>0$, show $$\prod_{c < p \leq x} \left(1 - \frac{c}{p} \right) \ll \log ^{-c} x ,$$ where $p$ is a prime.

I am not sure how to show this result. We know that $$\sum_{p \leq x} \log \left( 1-\frac{1}{p} \right) = -\log \log x - B + O\left (\frac{1}{\log x} \right)$$ for some constant $B>0$, but I am not sure if this helps much as the RHS does not seem to give a hint of the RHS of what we want to prove.

Compact
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  • You may easily compare $\left(1-\frac{c}{p}\right)$ with $e^{-c/p}$ and exploit Mertens' theorem(s) on $\sum_{p\leq x}\frac{1}{p}$. – Jack D'Aurizio Mar 15 '18 at 20:11
  • @JackD'Aurizio Could you elaborate on how I would use Mertens' Theorems on the sum you mentioned to obtain the result? I have trouble seeing how we can obtain the arbitrary power of a logarithm. (I tried reading https://math.stackexchange.com/questions/1039831/what-are-the-connections-between-the-three-mertens-theorem for reference on Mertens' theorems.) – Compact Mar 15 '18 at 21:05

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Note that $$\frac{t}{1+t} < \log (1+t) < t$$ for $t> -1$ and $t \neq 0$. Then \begin{align*} c\log \left( 1 - \frac{1}{p} \right) & > \frac{-c/p}{1 - 1/p} \\ & = \frac{p}{p-1} \left(- \frac{c}{p} \right) \\ & > - \frac{c}{p} \\ & > \log \left(1 - \frac{c}{p} \right) \end{align*} for $p >c$. Exponentiate both sides to obtain $$\left( 1 - \frac{c}{p} \right) < \left( 1 - \frac{1}{p} \right) ^c.$$ So, \begin{align*} \prod_{c<p<x} \left(1 - \frac{c}{p} \right) & < \prod_{c<p<x} \left(1 - \frac{1}{p} \right) ^c \\ & \ll \prod_{p<x} \\ & \ll \log ^{-c} x \end{align*}

Dominated Convergence Theorem
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