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While answering another question in MSE, I had used the following result which I thought was a trivial consequence of the prime number theorem and equidistribution. However, I realized from the comments that many people thought that this was not either true or counter intuitive. Hence I am posting this as a question looking for a proof or disproof.

Let $p_k$ be the $k$-th prime and $f$ be a continuous function Riemann integrable in $(0,1)$ such that

$$\lim_{n \to \infty}\frac{1}{n}\sum_{r = 1}^{n}f\Big(\frac{r}{n}\Big) = \int_{0}^{1}f(x)dx. $$

Then, $$ \lim_{n \to \infty}\frac{1}{n}\sum_{r = 1}^{n}f\Big(\frac{p_r}{p_n}\Big) = \int_{0}^{1}f(x)dx. $$

My proof was based on showing that as $n \to \infty$, the ratios $p_r/p_n$ approached equidistribution in $(0,1)$ hence the integral follows as a trivial property of equidistributed sequence.

Motivation: There are several identities, limits etc on prime numbers which can be easily proven using this simple formula, including all answers to all three questions on the arithmetic, geometric and harmonic means of primes mentioned in the above link.

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    $\begingroup$ If you take the example $f(x)=2x$, then the right-hand side is just one. However, for a huge (fixed) value of $n$, there would be more primes in the interval $[0,p_n/2]$ than in the interval $[p_n/2,p_n]$, so why whould the value of that average be near the number one? ADDITION: As an example, among the first half billion numbers there are 26355867 primes, but among the next half billion numbers, i.e. between $0.5\cdot 10^9$ and $10^9$, there are only 24491667 primes. $\endgroup$ – Jeppe Stig Nielsen Sep 18 '18 at 15:00
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    $\begingroup$ @JeppeStigNielsen How do you know that the first interval will have more primes asymptotically? $\endgroup$ – Yanko Sep 18 '18 at 15:03
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    $\begingroup$ @JeppeStigNielsen I think the same as you. For me, it would look nicer, if one changes $dx$ by another integrator (kinda Riemann-Stieltjes) related to the distribution of primes. $\endgroup$ – DCao Sep 18 '18 at 15:10
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    $\begingroup$ @JeppeStigNielsen use the prime number theorem: $\pi(n)/\pi(2n)\approx (n/\log (n))/(2n\log(2n)) = 1/2 (\log(2)+\log(n))/\log(n)\to 1/2$ $\endgroup$ – Bananach Sep 18 '18 at 15:13
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    $\begingroup$ @Marcuswood the prime number theorem says that F(x)=x indeed. This is essentially because the "average gap" $\log x$ between prime numbers is a slowly varying function, too slow to matter when the numbers of prime numbers at $n$ and $x n$ are considered with $n\to\infty$. $\endgroup$ – Bananach Sep 18 '18 at 15:24
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Too long for a comment

Divide $[0,1]$ into $$[0,p_1/p_n],[p_1/p_n,p_2/p_n],\cdots,[p_{n-1}/p_n,1]$$

Then, using Riemann sum, we have $$I:=\int^1_0f(x)dx=\lim_{n\to\infty}\sum^n_{k=1}f\left(\frac{p_k}{p_n}\right)\frac{p_{k+1}-p_k}{p_n}$$

If we assume that $p_j=j\ln j$, $$I=\lim_{n\to\infty}\sum^n_{k=1}f\left(\frac{p_k}{p_n}\right)h(k,n)+\lim_{n\to\infty}\frac1n \sum^n_{k=1}f\left(\frac{p_k}{p_n}\right) \qquad{(1)}$$ where $$h(k,n)=\frac{(k+1)\ln(k+1)-k\ln k}{n\ln n}-\frac1n$$

It can be shown that $$h(k,n)\le h(n,n)=O(\frac1{n\ln n})$$

Therefore, the absolute value of the first term in $(1)$ is upper bounded by $$h(n,n)\cdot nM\to 0$$ where $M$ is a positive constant. This leads us to our desired result.

I am not sure if this argument can be made rigorous. I will review it when I have leisure time.

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  • $\begingroup$ This is already an answer in with an approach different from mine. Waiting for your leisure time lol $\endgroup$ – Nilotpal Kanti Sinha Sep 20 '18 at 3:48
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Posting this as an answer rather than a comment because it contains the actual answer. Since I did not get a conclusive answer in MSE, I posted the question in MO where a rigorous proof was provided.

https://mathoverflow.net/questions/311085/calculating-limits-using-integration-for-sequence-of-prime-numbers

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