Taking the <natural logarithm> of the finite <Euler product> and using the <Taylor series>
$$
-\log(1-u)=u+\sum_{k\geq2}\frac{u^k}{k}
$$
gives
$$
\log\prod_{p\leq x}\left(1-\frac1p\right)^{-1}
=\sum_{p\leq x}\frac1p
+\sum_{p\leq x}\sum_{k\geq2}\frac1{kp^k}.
$$
The double series converges absolutely, and the <Mertens second theorem> therefore makes the right side
$$
\log\log x+\log C+O\left(\frac1{\log x}\right)
$$
for
$$
C=\exp\left(c+\sum_p\sum_{k\geq2}\frac1{kp^k}\right)>0.
$$
Exponentiating proves
$$
\boxed{\prod_{p\leq x}\left(1-\frac1p\right)^{-1}
=C\log x+O(1).}
$$
This is the <Mertens third theorem>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-150/1/d/solution