Is this a valid attempt at the Riemann Hypothesis?

Question

From Marcus Du Sautoy's book “The music of the primes”, there is a method of finding a very long list of N consecutive numbers which are not primes. e.g $101!+2, 101!+3,...,101!+101$ all of which are not primes $(N=100)$. In general $(N+1)!+2$ to $(N+1)!+N+1$ are not primes.

This means $\pi((N+1)!+1)$ to $\pi((N+1)!+N+1)$ is a flat line on the graph of $\pi(x)$ for however long N is.

The main arguments for this attempt at the Riemann Hypothesis is based on

1. An equivalent to the Riemann Hypothesis which states that $$|\text{Li}(x)-\pi(x)|\le c\sqrt{x}\ln x$$ i.e. $Li(x)$ is a good approximation of $\pi(x)$ *1
2. The idea that if Li($x$) is a bad approximation of $\pi(x)$ then when an unimaginably long extended flat line of $\pi(x)$ occurs then Li($x$) should at one point cease to be a good approximation.

Going forward let $\beta$ = (N+1)! and it follows that $\beta + 1$ is the start of our flat line

We would first consider $$\text{Li}(\beta+1) > \pi(\beta+1).$$ {1} Drawing a sketch yourself would most likely help to understand the following steps. We hope to show that $$\text{Li}(\beta+n) - \pi(\beta+n) \le c\sqrt{\beta+n}\ln(\beta+n) (1) $$ where $n$ is any integer $\le N+1$

We would also now assume that $$\text{Li}(\beta+1) - \pi(\beta+1) \le c\sqrt{\beta+1}\ln(\beta+1)$$ i.e. the beginning of our flat line and so this is a reasonable assumption as we are saying the Riemann hypothesis is true until now. Then consider and assume (like in induction) $$\text{Li}(\beta+k) - \pi(\beta+k) \le c\sqrt{\beta+k}\ln(\beta+k) (2)$$ For $n = k+1 $ $$\text{Li}(\beta+k+1) - \pi(\beta+k+1) = \text{Li}(\beta+k) - \pi(\beta+k) +(\text{Li}(\beta+k+1) - \text{Li}(\beta+k)) (3) $$note: $\pi(\beta+k+1) = \pi(\beta+k)$ as we are still in the flat line region.

Remembering the $\beta =(N+1)!$ If we make N large enough $$\text{Li}(\beta +k+1)-\text{Li}(\beta+k) \approx 0$$ Therefore it is safe to say that $$\text{Li}(\beta+k+1) - \pi(\beta+k+1) \leq c\sqrt{\beta+k+1}\ln(\beta+k+1)$$The worst case is that $\text{Li}(\beta+k) - \pi(\beta+k) = c\sqrt{\beta+k}ln(\beta+k)$ which using (3) which implies $$\text{Li}(\beta+k+1) - \text{Li}(\beta+k) \leq c\sqrt{\beta+k+1}\ln(\beta+k+1) - c\sqrt{\beta+k}\ln(\beta+k)$$ However both sides tend to zero for large $N$

So it is perhaps fair to conclude that if (1) is true for $n=k+1$ it is also true for n=k+1 and as we started with saying it was true for $n=1$ it must also be true for $n \leq N+1$

If $$\text{Li}(\beta+1) = \pi(\beta+1) $$ {2} then the argument is the same as above

However what if $$\pi(\beta+1) > \text{Li}(\beta+1) $$ {3} While as we can make $N$ as large as we want then we can be sure Li($x$) would eventually cross $\pi(x)$ and then we can a apply the argument for both {1} and {2} from the point where they cross

Eventually going back to our main argument 2 above we see that even in the most likely condition for Li($x$) to be a bad approximation it doesn't do so, which implies Li($x$) is a good approximation of $\pi(x)$

*1 Weisstein, Eric W. "Prime Number Theorem." From MathWorld--A Wolfram Web Resource.

Answer 1

It would theoretically be possible to disprove the Riemann Hypothesis by establishing the existence of a 'too long' string of composite numbers.

However, this 'too long' needs to be taken in a relative sense, that is relative to the size of the elements in the string. And, in your example the elements are huge relative to the length of the string.

Answer 2

Roughly speaking, according to the Prime Number Theorem, in the region of $x$, on average one integer out of every $\log x$ is prime. Now, Stirling's approximation says that $\log n!$ is roughly $n\log n$; so in the region of $n!$, you would expect one in every $n\log n$ numbers to be prime. This is less than one in every $n$.

So to find a run of composite integers of length $n$ in the region of $n!$ is nothing special; in fact, we would expect to find runs of length $n \log n$ and more without having to look hard.