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Let $ f: \mathbb{N} \to \mathbb{N} $ be a number-theoretic function satisfying $ f(xy) = f(x) + f(y) $ whenever $ \gcd(x,y) = 1 $.

How can I prove that $ \sum_{\substack{p ~ \text{prime}; \\ p \leq n}} f(p) \frac{p}{n} \bigg\lceil \frac{n}{p} \bigg\rceil \left\{ \frac{n}{p} \right\} \sim \frac{1}{2} \sum_{\substack{p ~ \text{prime}; \\ p \leq n}} f(p), $ i.e., the ratio of these two sums, as $ n $ tends to $ \infty $, is equal to $ 1 $?

Notation: $ \lceil \cdot \rceil $ denotes the ceiling function, and $ \{ \cdot \} $ the fractional-part function.

If someone could just simplify the problem, not even necessarily prove it, I would greatly appreciate it.

Also, I don't know if the following relations might help in a proof/simplification: \begin{align} \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} \bigg\lceil \frac{n}{k} \bigg\rceil \left\{ \frac{n}{k} \right\} &= \frac{1}{2}, \\ \lim_{n \to \infty} \frac{1}{n} \sum_{\substack{p ~ \text{prime}; \\ p \leq n}} \frac{p}{n} \bigg\lceil \frac{n}{p} \bigg\rceil \left\{ \frac{n}{p} \right\} &= \frac{1}{2}. \end{align}

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    (1) You wrote 'Also, idk if the following relations might help...', so I replaced 'idk' by 'I don't know'. (2) You asked how you could prove that the ratio is $ 1 $ and then tried to explain the meaning of $ \lceil \cdot \rceil $ and $ \{ \cdot \} $, all in one sentence! Such practice is not part of good mathematical exposition. Try to separate your question from comments about notation that you have used. Readers aren't in a rush to read your post, so you don't have to worry if they are unfamiliar with notation that you intend to explain right after the question is posed.2012-12-27

1 Answers 1

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The statement that $\sum_{\substack{p ~ \text{prime}; \\ p \leq n}} f(p) \frac{p}{n} \bigg\lceil \frac{n}{p} \bigg\rceil \left\{ \frac{n}{p} \right\} \sim \frac{1}{2} \sum_{\substack{p ~ \text{prime}; \\ p \leq n}} f(p)$ is not true in general.

Consider the following counterexample: \begin{equation} f(x)= \begin{cases} 1& \text{if $n$ is even} \\ 0& \text{if $n$ is odd} \end{cases} \end{equation}

Then $\frac{1}{2} \sum_{\substack{p ~ \text{prime}; \\ p \leq n}} f(p)=\frac{1}{2}$ and \begin{equation} \sum_{\substack{p ~ \text{prime}; \\ p \leq n}} f(p) \frac{p}{n} \bigg\lceil \frac{n}{p} \bigg\rceil \left\{ \frac{n}{p} \right\}= \frac{2}{n} \bigg\lceil \frac{n}{2} \bigg\rceil \left\{\frac{n}{2}\right \} \begin{cases} 0& \text{if $n$ is even} \\ \frac{n+1}{2n} & \text{if $n$ is odd}\end{cases} \end{equation}

Thus the ratio of the 2 sums has no limit. (Though if we restrict $n$ to be odd then the limit of the ratio is indeed 1)

Are there additional restrictions?