{c}

There is a constant $C>0$ such that
$$
\prod_{p\leq x}\left(1-\frac1p\right)^{-1}
=C\log x+O(1).
$$
Taking logarithms reduces the result to the <Mertens second theorem>, because the terms of order $p^{-2}$ and smaller form an absolutely convergent series.

Mertens third theorem

Put
$$
A(x)=\sum_{p\leq x}\frac{\log p}{p}=\log x+E(x),
\qquad E(x)=O(1).
$$
The <Abel summation formula> with weight $1/\log t$ gives
$$
\sum_{p\leq x}\frac1p
=\frac{A(x)}{\log x}
+\int_2^x\frac{A(t)}{t(\log t)^2}\,dt.
$$
Substituting $A(t)=\log t+E(t)$ yields
$$
\sum_{p\leq x}\frac1p
=\log\log x+c+\frac{E(x)}{\log x}
-\int_x^\infty\frac{E(t)}{t(\log t)^2}\,dt
$$
for a constant $c$. Both final terms are $O(1/\log x)$, so
$$
\boxed{\sum_{p\leq x}\frac1p
=\log\log x+c+O\left(\frac1{\log x}\right).}
$$
This is the <Mertens second theorem>.

Solution

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

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>.

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

{c}

= Mertens theorem for reciprocal primes
{c}
{synonym}

There is an absolute constant $B_1$, the Meissel–Mertens constant, such that
$$
\sum_{p\leq x}\frac1p
=\log\log x+B_1+O\left(\frac1{\log x}\right).
$$


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 Incoming links: Mertens second theorem