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Show that $$\sum_{p \leq x} \log \left( 1-\frac{1}{p} \right) = -\log \log x - B + O\left (\frac{1}{\log x} \right)$$ and deduce $$\prod_{p\leq x} \left(1-\frac{1}{p} \right) = \frac{e^{-B}}{\log x} + O\left( \frac{1}{\log ^2 x } \right)$$

So far I have not had luck in showing either equality. I have proved the following: $$\sum_{p\leq x} \frac{\log p}{p} = \log x + O(1),$$ $$\sum_{p\leq x} \frac{1}{p} = \log \log x + C + O\left( \frac{1}{\log x} \right).$$ With these two, I am not sure how to combine them to obtain $\log \left( 1-\frac{1}{p} \right)$, rather than $\frac{( \log p) - 1}{p}$. Any ideas on how to proceed/conclude with both equalities?

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Observe we have that \begin{align} \log\left(1-\frac{1}{p}\right) = -\frac{1}{p}+\mathcal{O}\left(\frac{1}{p^2}\right) \end{align} then we see that \begin{align} \sum_{p \leq x}\log\left(1-\frac{1}{p}\right) = -\sum_{p\leq x}\frac{1}{p}+\sum_{p\leq x}\mathcal{O}\left(\frac{1}{p^2}\right) = -\sum_{p\leq x}\frac{1}{p} + \text{const}. \end{align} Now, use what you have proved.

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