Upper bound for $\prod_{p\leq x}\Big(1-\frac{1}{p}\Big)^{-1}$

Question

In an analytic number theory textbook (by Jörg Brüdern) I found this exercise:

Prove $$ \prod_{p \leq x}\left(1-\frac{1}{p}\right)^{-1}=B \log x+O(1) $$ with a fitting $B\in\mathbb{R}$.

Obviously I can write $$\prod_{p\leq x}\Big(1-\frac{1}{p}\Big)^{-1}=\sum_{n\leq x}\frac{1}{n} + \sum_{n>x, p\mid n\Rightarrow p\leq x} \frac{1}{n}$$ with $p$ being prime. I showed that $$\sum_{n\leq x}\frac{1}{n} \leq 1+ \log(x) $$ and spend quite much time trying to find any upper bound for the second sum. It had various ideas but none of them worked out.

Then I got the idea to consider the product

$$\prod_{p\leq x} (1+\frac{1}{p}):(1-\frac{1}{p})^{-1}=\prod_{p\leq x} (1+\frac{1}{p})(1-\frac{1}{p})= \prod_{p\leq x} (1-\frac{1}{p^2})=O(1)$$ because this last product is converging for $x\to\infty$ (because the sum $\sum_{n=1}^\infty\vert{n^{-2}}\vert$ is converging, right?)

Since I already proved (in another exercise) that $$\prod_{p\leq x} (1+\frac{1}{p})=O(\log(x))$$

it immediately follows that $$\prod_{p\leq x}(1-\frac{1}{p})^{-1}= B\log(x)+O(1)=O(\log(x))$$

because the quotient is $O(1)$.

So my question is, is this above a valid argument to proof the statement?

And is there any "direct" way to estimate the right sum above? If so, I'd appreciate any hints on this.

Brüdern, Jörg (2013). Einführung in die analytische Zahlentheorie. Springer-Verlag.

Answer 1

$$ \log\prod_{p\leqslant x}\left(1-\frac{1}{p}\right)=-\sum_{p\leqslant x}\frac{1}{p}+\sum_{p\leqslant x}\left(\log\left(1-\frac{1}{p}\right)+\frac{1}{p}\right) $$ The second sum converges because the general term is $\mathcal{O}\left(\frac{1}{p^2}\right)$, as for the first sum we have the estimation $\sum_{p\leqslant x}\frac{1}{p}=\log\log x+c+\mathcal{O}\left(\frac{1}{\log x}\right)$ for a constant $c$. Therefore, using the fact that the rest of the series $\sum\frac{1}{n^2}$ is a $\mathcal{O}\left(\frac{1}{n}\right)$, we have $$ \log\prod_{p\leqslant x}\left(1-\frac{1}{p}\right)=-\log\log x+A+\mathcal{O}\left(\frac{1}{\log x}\right) $$ for a constant $A$. Finally, $$ \prod_{p\leqslant x}\left(1-\frac{1}{p}\right)^{-1}=e^A \log x+\mathcal{O}(1) $$

Answer 2

This answer aims to determine the constant in the result proven in @Tuvasbien's answer:

$$ \prod_{p\le x}\left(1-\frac1p\right)^{-1}=e^\gamma\log x+O(1). $$

where $\gamma$ is the Euler-Mascheroni constant. As @Tuvasbien has pointed out in his answer, it suffices to show that

$$ S(x)=\sum_{p\le x}\log\left(1-\frac1p\right)^{-1}=\log\log x+\gamma+O\left(1\over\log x\right). $$

In fact, when $s\to1$ from the right, the function $\zeta(s)$ satisfies the identity:

$$ \zeta(s)=\prod_p\left(1-{1\over p^s}\right)^{-1}\sim{1\over s-1}. $$

Taking logarithm on both side indicates that as $\delta\to0$, there is

\begin{aligned} \log\zeta(1+\delta) &=\log\frac1\delta+O(\delta)=\log{1\over1-e^{-\delta}}+O(\delta) \\ &=\sum_{n\ge1}{e^{-n\delta}\over n}+O(\delta). \end{aligned}

Let $H(x)$ denote the sum of reciprocal of positive integers $\le x$:

$$ H(x)=\sum_{n\le x}\frac1n=\log x+\gamma+O\left(\frac1x\right). $$

Then the above expression gets to transformed into an integral:

$$ \log\zeta(1+\delta)=\delta\int_0^\infty e^{-\delta t}H(t)\mathrm dt+O(\delta)\tag1 $$

For convenience, we introduce the following arithmetical function $a_n$:

$$ a_n= \begin{cases} 1/kp^k & n=p^k,p\text{ prime} \\ 0 & \text{otherwise} \end{cases} $$

so that $\log\zeta(s)$ can be transformed into

$$ \log\zeta(1+\delta)=\sum_p\sum_{k\ge1}{1\over kp^{k(1+\delta)}}=\sum_{n\ge1}{a_n\over n^\delta}, $$

and $S(x)$ can become

$$ S(x)=\sum_{p\le x}\sum_{k\ge1}a_{p^k}=\sum_{n\le x}a_n+R(x), $$

where $R(x)$ satisfies

\begin{aligned} |R(x)| &=\sum_{p\le x}\sum_{k>\log_px}a_{p^k}\le\sum_{p\le x}\sum_{k>\log_px}{1\over p^k} \\ &=\sum_{p\le x}\sum_{m\ge0}{1\over p^{m+\lceil\log_p x\rceil}}\le\frac1x\sum_{p\le x}{1\over1-p^{-1}} \\ &\le\frac1x\sum_{p\le x}2={2\pi(x)\over x}=O\left(1\over\log x\right). \end{aligned}

Consequently, it follows from @Tuvasbien's answer that

$$ S_2(x)=\sum_{n\le x}a_n=\log\log x+A+O\left(1\over\log x\right). $$

Now, using the fact that

\begin{aligned} \log\zeta(1+\delta) &=\sum_{n\ge1}{a_n\over n^\delta}=\delta\int_1^\infty x^{-\delta}S_2(x){\mathrm dx\over x} \\ &=\delta\int_0^\infty e^{-\delta t}S_2(e^t)\mathrm dt \end{aligned}

Combining this with (1) gives

\begin{aligned} \log\zeta(1+\delta)-\log{1\over1-e^{-\delta}} &=\delta\int_0^\infty e^{-\delta t}[S_2(t)-H(t)]\mathrm dt \\ &=A-\gamma+O\left\{\delta\int_0^\infty{e^{-\delta t}\over t+1}\mathrm dt\right\} \end{aligned}

Finally, using the fact that

\begin{aligned} \int_0^\infty{e^{-\delta t}\over t+1}\mathrm dt &=\int_0^T{e^{-\delta t}\over t+1}\mathrm dt+O\left\{\int_T^\infty{e^{-\delta t}\over t}\mathrm dt\right\} \\ &=O(T)+O\left\{\int_{\delta T}{e^{-u}\over u}\mathrm du\right\} \\ &=O(T)+O\left(1\over\delta T\right). \end{aligned}

Plugging this back with $T=\delta^{-1/2}$ gives us

$$ \log\zeta(1+\delta)-\log{1\over1-e^{-\delta}}=A-\gamma+O(\delta^{1/2}) $$

Since the left hand side is $O(\delta)$ when $\delta\to0$, we conclude that $A=\gamma$, thereby proving the identity stated in the front of this answer.