Asymptotic for sum of $(p-1)/p$ for $p$ at most $x$

Question

I found an asymptotic for $\sum\limits_{p\leq x}\dfrac{p-1}{p}$ where the sum is extended over primes not exceeding $x$ in an article but I cannot find it now. I can find the following using Abel partial summation (assuming my calculation is correct) $$\dfrac{x}{2\log{x}}-x+O\!\left(\dfrac{x}{\log^{2}{x}}\right)$$ but I think the bound I saw was sharper. Does anybody know a sharper bound for this? This is probably a duplicate but the search function does not really work well for mathematical terms and the expression is not easy to express verbally.

Answer 1

Your sum is $$ \sum\limits_{p \le x} {\frac{p-1}{p}}= \pi (x) - \sum\limits_{p \le x} {\frac{1}{p}} . $$ The sum on the right-hand side is $\sim \log\log x$ and any error term of any reasonable asymptotics for $\pi(x)$ is larger in size than that. Thus, your original sum will have the same asymptotic approximations as $\pi(x)$. Hence, for example, $$ \sum\limits_{p \le x} {\frac{p-1}{p}}= \int_2^x {\frac{{dt}}{{\log t}}} + \mathcal{O}(xe^{ - c\sqrt {\log x} } ). $$ Under the Riemann hypothesis, the error term can be improved to $\mathcal{O}(\sqrt{x}\log x)$.

Answer 2

It is equivalent to $$\pi(x)-\sum_{p\le x}\frac{1}{p}$$ It's well known that $\sum_{p\le x}\frac{1}{p}=\log(\log(x))+O(1)+O(1/\log^2(x))$, see Does the sum of reciprocals of primes converge?.

It's also well known that $\pi(x)=\frac{x}{\log(x)}\left(1+O(1/\log x)\right)$.

Hence the entire sum is $$\frac{x}{\log(x)}\left(1+O(1/\log x)\right)-\log\log x-O(1)-O(1/\log^2(x))$$