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: I shall bring corrections soon.

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

(The file content is that of Version 100 with a little justification added in Part 2 regarding the $\forall k \in \{0,\ldots,n-2\}, \Delta_{n-k} \ge 0 $ condition)

I HAVE CORRECTIONS TO DO HERE AS WELL. I'll edit this description whenever possible, unfortunately I can't right now while I publish this new version.

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:

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

$$\sum_{k=1}^{n} x_k^2 + 2r_n\sum_{k=1}^{n-1} \sum_{j=k+1}^{n} x_kx_j \xrightarrow[n \to \infty]{} 0, \space with \space x_1,\ldots,x_n \in \mathbb{R} \space and \space x_1=1,$$

The fact that all the $x_k$ are real requires all the corresponding $\Delta_k$ to be positive as well, which then requires:

$$\Rightarrow r_n=-\frac{1}{n+\alpha} \space with \space \alpha \in ]-3,-2[ \cup ]-2,-1]$$ (actually -3 has to be excluded I didn't pay attention to it in the paper sorry)

Then we shall substitute the $x_k$ to come back to our problem:

$$\forall k \in \{1,\ldots,n\}, \space x_k=\frac{1}{k^{a_n}}$$

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 $\frac{1}{2}$


Then I conclude that the only logically consistent case is when $a_n$ converges to $\frac{1}{2}$.

Actually, I find that the "real part"'s asymptotic equivalent has to be, as $n \to \infty$: $$\frac{1}{2} + o\left(\frac{1}{\ln(n)^2}\right) + ik\pi\left(\frac{\ln(n)+4-\lambda}{\ln(n)^2} + o\left(\frac{1}{\ln(n)^2}\right)\right), \space k \in \mathbb{Z}^*, \space \lambda \in [0,1]$$ and that it is therefore in fact the imaginary part that has a slower (so more prominent) rate of convergence $ik\pi\left(\frac{\ln(n)+4-\lambda}{\ln(n)^2}\right), \space k \in \mathbb{Z}^*, \space \lambda \in [0,1]$.

And this imaginary rate of convergence in that pseudo-real-part vanishes as $n \to \infty$, yielding $\frac{1}{2}$

EDIT:

- Several versions ago, My concern was that I had figured out $\sum_{k=1}^{n} \frac{1}{k^{2a_n}} - \frac{1}{n}\left(\sum_{k=1}^{n} \frac{1}{k^{a_n}}\right)^2 \xrightarrow[n \to \infty]{} 0$ precisely because I deemed all the $x_k$ as reals, which required that $r_n=-\frac{1}{n+\alpha} \space with \space \alpha \in [-3,-2[ \cup ]-2,-1]$, but that in the end, in the expression of its asymptotic equivalent as $n \to \infty$ (and NOT for $n$ sufficiently large as I write in the end of the document, sorry), $\frac{1}{k^{a_n}}$ ended up a complex, especially for the $k$ that are close to $n$, seemingly loosening the "all $\Delta_{n-k} \ge 0$" condition which in turn would abolish that constraint on $r_n$.But in fact we're talking as $n \to +\infty$, for all the $\frac{1}{k^{a_n}}$, $n$ has already tended to infinity so we actually end up with $\frac{1}{\sqrt{k}}$ which are indeed real, sooner or later all the $k$ are "left infinitely far behind" by $n$ as it tends to infinity.What this means is simply that there are no real solutions in the form $\frac{1}{k^{a_n}}$ for any finite $n$ to the equation $\sum_{k=1}^{n} \frac{1}{k^{2a_n}} - \frac{1}{n}\left( \sum_{k=1}^{n} \frac{1}{k^{a_n}} \right)^2 = 0$.But everything goes well when we fly to infinity and beyond.

- More recently, My doubts about the quadratic equations were actually exaggerated, those led me to write a completely erroneous version in which I made weird statements. I thought about a case where a $\Delta_{n-k} \ge 0$ condition wouldn't be satisfied, removing the constraint on $r_n$ on which we rely for the comparison of the convergence cases, but it would precisely mean that the corresponding $x_{n-k}$ is a complex, which would contradict all the $x_k$ belonging to $\mathbb{R}$.So in the end all the $\Delta_{n-k}$ are positive and we are safe.This was really odd given that my previous concern precisely was about the necessity of having all the $x_{n-k} $ real to require $\Delta_{n-k} \ge 0$.I admit that I've been a little goofy on this. I think I am really tired.

The reasons why I have so many doubts and "dumb" questionings is because I want to rigorously stress test my statements. And sometimes it's just over-exaggerated like this panick attack regarding the $\Delta_{n-k}$.

NOTES:

This version has the same content as Version 100 with a little sentence added in the quadratic equations part to justify all that the $\Delta_{n-k}$ must be positive.

I'm not far from the final version (I hope).

Feedback would be appreciated at: emrekaplan1978@gmail.com