A Possible Proof Of The Riemann Hypothesis
Creators: name: KAPLAN, Yunus-Emre
Publisher: name: Zenodo
Year: 2025
Subjects: Riemann Hypothesis
Asymptotic Analysis
Prime Numbers
Imbricated Quadratic Equations
Integrals
Zeta Function
Eta Function
Taylor Expansion

description: PLEASE DON'T DOWNLOAD THIS,GO TO THE LAST VERSION INSTEADLast version as I write this: 104

My last version when I write this is 103, Version 100 is the right one.

Please refer to Version 100. The constraints I thought were not satisfied were actually satisfied. All the x_k must be real so all the $\Delta_{n-k}$ must be positive.

This study is my approach to the Riemann Hypothesis, which potentially provides elements for a proof of it, or even an actual proof.

Before starting I define $s_n=a_n+ib_n$, a zero to the $n$-th order truncation of $\eta$, $\eta_n$.

I start by breaking down the equality $\eta(s)=0$ (which is implied by the nullity of the continued $\zeta(s)$) by expressing $|\eta(s)|^2=0$ as $\Re(\eta(s))^2 + \Im(\eta(s))^2 = 0$, expressed with cosine and sine functions, then I use trigonometric identities to further break down the sum into a sum of squares purely relying on $a_n=\Re(s_n)$ and a double sum relying on both $a_n=\Re(s_n)$ and $b_n=\Im(s_n)$ where $s_n$ is a zero of the truncated $\eta_n(s_n)$:

$$\sum_{k=1}^{n} \frac{1}{k^{2a_n}} + 2\sum_{k=1}^{n-1} \sum_{j=k+1}^{n} \frac{(-1)^{k+j-2}\cos(b_n\ln(k/j))}{(kj)^{a_n}} \xrightarrow[n \to \infty]{} 0$$

The emergence of the sum of squares $\sum_{k=1}^{n} \frac{1}{k^{2a_n}}$ is what shows us we're on the right path, for we can compare it with $\frac{n^{1-2a_n}-1}{1-2a_n}$, and since we want to prove $\Re(s_n)=a_n\xrightarrow[n \to +\infty]{}\frac{1}{2}$, that's a very good sign to see $1-2a_n$ appear.

Using the bounded nature of the cosine function, I reformulate the equality as a problem of quadratic equations, to get rid of the dependence on $\Im(s_n)$ as well as the oscillations it induces:

$\forall x \in \mathbb{R}, -1 \le \cos(x) \le 1$ so for all $n$ sufficiently large, $ \space \exists r_n \in \mathbb{N} \space | \space -1 \le r_n \le 1$ so that:

$$ \sum_{k=1}^{n} \frac{1}{k^{2a_n}} + 2r_n\sum_{k=1}^{n-1} \sum_{j=k+1}^{n} \frac{1}{(kj)^{a_n}} \xrightarrow[n \to \infty]{} 0$$

It has already been proved that for all the nontrivial zeros, their real part is in $]0,1[$, therefore, using the $a_n \xrightarrow[n \to \infty]{} a_{+\infty} \in ]0,1[$ condition and asymptotic equivalences, we end up with:

$$\sum_{k=1}^{n} \frac{1}{k^{2a_n}} - \frac{1}{n}\left( \sum_{k=1}^{n} \frac{1}{k^{a_n}} \right)^2 \xrightarrow[n \to \infty]{} 0$$

I then use integrals and asymptotic analysis to observe three cases of asymptotic behaviours:



when $a_n$ converges to a real number in $]0,\frac{1}{2}[$

when $a_n$ converges to a real number in$ ]\frac{1}{2},1[$

when $a_n$ converges to $\frac{1}{2}$


So far we only lack proving that the case when $a_n$ converges to a real number in$ ]\frac{1}{2},1[$ doesn't allow any adequate rate of convergence ensuring the limit.

NOTES:

After a thorough analysis of the necessary conditions in the part with the quadratic equations and the calculation of all the $\Delta_{n-k}$, I finally concluded that that part was nearly useless, and brings no more actual constraint to $r_n$ than expressing it as a fraction of the two sums.

I could indeed prove (if I made no error) that the real part could be $\frac{1}{2}$, couldn't be in $]0,\frac{1}{2}[$, but couldn't find an interesting constraint for the $]\frac{1}{2},1[$ case.

I am really sorry if I failed you. At least we discarded the first half of the $]0,1[$ interval and could confirm for $\frac{1}{2}$.