Mertens' third theorem for primes

By [L2], the sum in [L1] is exactly the finite prime-power sum $S(x):={\sum }_{p^k\le x}\frac{1}{k{p}^k}\mathrm{.}$ Hence $S(x)=\log \log x+\gamma +O(1\mathrm{/}\log x)\mathrm{.}$

Every factor $1-1\mathrm{/}p$ is positive. Applying [L3] with $u=1\mathrm{/}p$ and summing the resulting absolutely convergent series gives $-\log {\prod }_{p\le x}(1-\frac{1}{p})={\sum }_{p\le x}{\sum }_{k\ge 1}\frac{1}{k{p}^k}\mathrm{.}$

The difference between the sum in step 1.2 and $S(x)$ consists of terms with $k\ge 2$, $p\le x$, and $p^k>x$. For a fixed $k$, comparison with the positive decreasing series $\sum n^{-k}$ gives ${\sum }_{\begin{array}{c}p\le x\\ p^k>x\end{array}}\frac{1}{k{p}^k}=\{\begin{array}{ll}O(x^{-1\mathrm{/}2}),& 2\le k\le \lfloor {\log }_{2}x\rfloor ,\\ O(2^{-k}),& k>\lfloor {\log }_{2}x\rfloor \mathrm{.}\end{array}$ Indeed, in the first range $x^{1\mathrm{/}k}\ge 2$ and the integral tail is at most a constant times $x^{-(k-1)\mathrm{/}k}\le x^{-1\mathrm{/}2}$; in the second range the full tail from $n=2$ is $O(2^{-k})$. Summing over $k$ gives ${\sum }_{p\le x}{\sum }_{k\ge 1}\frac{1}{k{p}^k}=S(x)+O(\log x\mathrm{/}\sqrt{x})+O(1\mathrm{/}x)=S(x)+O(1\mathrm{/}\log x)\mathrm{.}$

Combining steps 1.1, 1.2, and 2.1 yields $\log {\prod }_{p\le x}(1-\frac{1}{p})=-\log \log x-\gamma +O(1\mathrm{/}\log x)\mathrm{.}$ Exponentiating the bounded $O(1\mathrm{/}\log x)$ term gives ${\prod }_{p\le x}(1-\frac{1}{p})=\frac{e^{-\gamma }}{\log x}(1+O(1\mathrm{/}\log x))\mathrm{.}$