Prime number sum Let $p$ denote a prime, and let $\{x\}$ denote the fractional part of $x$.
Suppose that the following statement is true for all non-integer real numbers $x$:
$$\lim_{n\to\infty}\frac{\sum_{p\leq n}^\ \frac{\ln(p)}{p}\{ px \}}{\sum_{p\leq n}^\ \frac{\ln(p)}{p}}=\frac{1}{2}.$$
Does that imply that
$$\lim_{n\to\infty}\frac{\sum_{p\leq n}^\ 1\cdot\{ px \}}{\sum_{p\leq n}^\ 1}=\frac{1}{2}?$$
 A: This is only a partial answer, but the conclusion can be proved (directly, without using the hypothesis) when $x$ is a rational number. Suppose $x=a/b$ in lowest terms is fixed. Then
$$
\sum_{p\le n} \bigg\{\!\frac{pa}b\!\bigg\} = \sum_{\substack{1\le c\le b \\ (c,b)=1}} \sum_{\substack{p\le n \\ pa\equiv c\!\pmod{\!b}}} \bigg\{\!\frac{pa}b\!\bigg\} + \sum_{p\mid b} \bigg\{\!\frac{pa}b\!\bigg\},
$$
because every prime $p$ either divides $b$ or is relatively prime to $b$, in which case $pa$ is relatively prime to $b$ as well. Since $\{t/b\}$ is periodic modulo $b$, this becomes
\begin{align*}
\sum_{p\le n} \bigg\{\!\frac{pa}b\!\bigg\} &= \sum_{\substack{1\le c\le b \\ (c,b)=1}} \bigg\{\!\frac{c}b\!\bigg\} \sum_{\substack{p\le n \\ pa\equiv c\!\pmod{\!b}}} 1 + O(1) \\
&= \sum_{\substack{1\le c\le b \\ (c,b)=1}} \frac{c}b \pi(n;b,c) + O(1),
\end{align*}
and so
$$
\lim_{n\to\infty} \frac{\sum_{p\le n} \big\{\!\frac{pa}b\!\big\}}{\sum_{p\le n} 1} = \lim_{n\to\infty} \sum_{\substack{1\le c\le b \\ (c,b)=1}} \frac{c}b \frac{\pi(n;b,c)}{\pi(n)}.
$$
The prime number theorem for arithmetic progressions tells us that $\pi(n;b,c)/\pi(n)$ tends to $1/\phi(b)$ when $n$ tends to infinity (since $c$ and $b$ are relatively prime); therefore
$$
\lim_{n\to\infty} \frac{\sum_{p\le n} \big\{\!\frac{pa}b\!\big\}}{\sum_{p\le n} 1} = \frac1{\phi(b)} \sum_{\substack{1\le c\le b \\ (c,b)=1}} \frac{c}b;
$$
this can easily be seen to equal $\frac12$ by pairing $c$ with $b-c$ in the sum (which works for all $b\ge3$).
