# asymptotic of a product

So the question that I'm working on is the following.

Show that $\Pi_{p\leq z}(1-\dfrac{1}{p})=\dfrac{C(1+\mathcal{o}(1))}{\log z}$.

First off I take logs and just work with the sum and thisis what I get. $$\sum_{p\leq z}\log (1-\dfrac{1}{p})=-\sum_{p\leq z}\sum_{n=1}^{\infty}\dfrac{1}{np^{n}}$$

now so now I want to show that this sum is $-\log\log z +C' +\mathcal{o}(\log x^{-1})$.

But I'm not sure on who to do this, anyone got any ideas?

• I show what you seek in this answer by appealing to a strengthened version of Mertens' formula, but I'm not sure if that's the road you'd like to take. – Antonio Vargas Jan 24 '14 at 15:09
• Actually, since you only want to show that $\prod_{p \leq z} (1-1/p) \sim C/\log z$ then you don't need the strengthened formula with the little o. Using the elementary version of Mertens' formula with error $O(1/\log z)$ will still yield the desired answer. – Antonio Vargas Jan 24 '14 at 16:30

## 1 Answer

$$\ln(n)=\sum_{d\mid n}\Lambda(d)$$ $$\ln(\lfloor x \rfloor!)=\sum_{n\leq x}\Lambda(n)\lfloor\frac{x}{n}\rfloor$$ $$x\ln(x)+O(x)=x\sum_{p\leq x}\frac{\ln(p)}{p}+O(x)$$ $$\sum_{p\leq x}\frac{\ln(p)}{p}=\ln(x)+O(1)$$ $$\text{ Now conclude your result using summation by parts}$$

• And use summation by parts on what sum? also I don't see how you can get the little oh out of this? – lance wellton Jan 24 '14 at 5:38
• Yea my bad, you wont be able to get the little O term out of this, I thought you were asking for something weaker. You could look at Tao's terrytao.wordpress.com/2013/12/11/mertens-theorems, if your comfortable with his Fourier analytic stuff. – Ethan Jan 24 '14 at 5:48