# Attempted exercise using Littlewood's theorem

This was an exercise to try to show we can use Littlewood's theorem$^1$ to prove that

$$\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^{N} \frac{g(n)}{\log p_n} = 1 \hspace{30mm}(1)$$

If $\vartheta(p_k) \approx p_k$ and $\vartheta(p_{k+m}) \approx p_{k+m}$ at two consecutive crossings $\vartheta(x) \approx x$ then number of prime gaps and primes is equal between crossings and $\sum \log p = \sum g(n)$ on the interval.

If we divide the interval between crossings into prime gaps on one hand and distances $\log p$ on the other, since the number of such sub-intervals and prime gaps is the same, the average must be close to 1 since (1) is basically a telescoping series.$^2$

At first I thought this would not work because we do not know that the crossings are far enough apart to assure the limit in (1), but we do not need to use consecutive crossings--we can choose them as far apart as we like.

Does this work? Thanks for any insights (especially about the telescoping series argument...).$^3$

1-Littlewood-Schmidt's theorem holds that $\vartheta(x) - x$ changes sign infinitely often. $\vartheta(x) = \sum_{p\leq x} \log p.$ I am using "crossing" to mean places where $\vartheta(x)$ crosses the line $y = x.$ I am using $g(n) = p_{n+1}- p_n,$ a prime gap.

2-My proof of the idea of the telescoping series was basically to show first that the average prime gap is $D/N$ in which D is the distance between crossings. By inverting the fraction we then get that the average value of $\log p$ is also $D/N.$ Can't we do this in one step?

3-The proof here is nicer because it does not use RH, which Littlewood does.

• Could you add the definition of $g$ also? Jul 31 '13 at 13:17
• Oh drats, I missed that, even though I was looking at that footnote. Never mind, then. Jul 31 '13 at 13:23

You can indeed use this approach to show that between any pair of crossings the average values of $\log p_n$ and $g(n)$ are close, and hence that $$\lim_{N\to\infty} \left(\frac{1}{N}\sum_{n=1}^N\log p_n-\frac{1}{N}\sum_{n=1}^N g(n)\right)=0 \\ \lim_{N\to\infty} \frac{\frac{1}{N}\sum g(n)}{\frac{1}{N}\sum \log p_n}=1$$
However, it doesn't follow that since the ratio of the averages goes to 1 that the average of the ratios also goes to 1. Something additional about the distribution of $\log p_n$ is needed to reach the conclusion.
In particular note that we can also write $$\lim_{N\to\infty} \frac{\frac{1}{N}\sum \log p_n}{\frac{1}{N}\sum g(n)}=1$$ but it does not seem to be true that $$\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^N \frac{\log p_n}{g(n)}=1$$