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 $$\lim_{n\to\infty}\frac{\sum_{p\leq n}^\ 1\cdot\{ px \}}{\sum_{p\leq n}^\ 1}=\frac{1}{2}?$$

  • $\begingroup$ The converse certainly holds, by partial summation. I believe that partial summation cannot prove the implication you want - in the same way we cannot derive the prime number theorem from Mertens's formula. $\endgroup$ – Greg Martin Dec 16 '12 at 22:31
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    $\begingroup$ On the other hand, your conclusion is true when $x$ is a rational number - it follows from the prime number theorem for arithmetic progressions, which tells us that the primes are distributed equally among the reduced resides modulo the denominator of $x$. $\endgroup$ – Greg Martin Dec 16 '12 at 22:33
  • $\begingroup$ Greg If I offer it up as a bounty will you give me a proof that the first one implys the second. I can already prove that the first statement is true. $\endgroup$ – Ethan Dec 17 '12 at 0:32
  • $\begingroup$ I wrote a partial answer below. I can't prove that the first implies the second - although I would be interested in seeing your proof of the first statement: one might be able to modify it to make a proof of the second statement. $\endgroup$ – Greg Martin Dec 17 '12 at 3:18
  • $\begingroup$ To answer the question directly: I don't believe so, or at least not without extra assumptions. I do not think that the first statement implies the second without something stronger. Going the other way however is a standard application of summation by parts. The reason is exactly the same as why Mertens estimate is weaker than the prime number theorem. That is, the statement $\sum_{p\leq n} \frac{\log p}{p} \sim \log n$ is strictly weaker than the PNT $\sum_{p\leq n} 1 \sim \frac{n}{\log n}.$ $\endgroup$ – Eric Naslund Dec 17 '12 at 5:35

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 $b\ge3$).

  • $\begingroup$ I appreciate the work you spent, but I was trying not to use dirichlets theorem $\endgroup$ – Ethan Dec 17 '12 at 4:32
  • $\begingroup$ I understand. I'd still be interested in seeing your proof of the first statement. $\endgroup$ – Greg Martin Dec 17 '12 at 7:37
  • $\begingroup$ Greg, I have done alot of work on this and some other related topics , can I send you some of my other work, and you give me advice on it? I can give the limiting difference of the first sum and 1/2ln(n), for all real numbers x, in terms of a complex polylogarithm. If you have an email, I would gladly send you some of my work, I wont include proofs unless you want them. None of the tools I use go beyond the use of several fourier series expansions, and some algebraic manipulation. $\endgroup$ – Ethan Dec 17 '12 at 18:01
  • $\begingroup$ You can reach me at gerg@math.ubc.ca (not a typo). $\endgroup$ – Greg Martin Dec 17 '12 at 20:01
  • $\begingroup$ I sent you 2 papers, id appreciate any advice $\endgroup$ – Ethan Dec 17 '12 at 20:22

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