A Possible Proof Of The Riemann Hypothesis

This study is my approach to the Riemann Hypothesis, involving the analysis of the convergence of the zeros of the partial sum $\eta_n$ of the Dirichlet Eta function $\eta$. Before starting I define $s_n=a_n+ib_n$, a zero to the squared modulus of the truncation of the $n$-th order of $\eta$, $|\eta_n(s_n)|^2=0$. 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 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: $$ \sum_{k=1}^{n} \frac{1}{k^{2a_n}} + 2r_n\sum_{k=1}^{n-1} \sum_{j=k+1}^{n} \frac{1}{(kj)^{a_n}} = 0$$ $$ \Leftrightarrow 2r_n\left(\sum_{k=1}^{n-1} \frac{1}{k^{a_n}}\right)\frac{1}{n^{a_n}} + \sum_{k=1}^{n} \frac{1}{k^{2a_n}} + 2r_n\sum_{k=1}^{n-2} \sum_{j=k+1}^{n-1} \frac{1}{(kj)^{a_n}} = 0$$ Using integrals and Taylor expansions, I end up with the following necessary condition: $$\lambda + \frac{n^{1-2a_n}-1}{1-2a_n}\xrightarrow[n \to \infty]{} 0$$$$with \space \lambda \in [0,1]$$ I then compare two cases of asymptotic behaviours: when $a_n$ converges to a real number in $\Big]\frac{1}{2},1\Big[$ when $a_n$ converges to $\frac{1}{2}$ without having to deal with the $\Big]0,\frac{1}{2}\Big[$ case, because the Riemann's functional equation $\zeta(s)=2^{s}\pi^{1-s}\sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s)$ implies that no zero for $Re(s)\in\Big]\frac{1}{2},1\Big[$ means no zero for $Re(s)\in\Big]0,\frac{1}{2}\Big[$ as well. Then I conclude that the only logically consistent case is when $a_n$ converges to $\frac{1}{2}$. By the Hurwitz's Theorem, we then conclude that given the uniform convergence of $\eta$ for $Re(s)\in[\sigma_1,\sigma_2], $ for any subset $[\sigma_1<\sigma_2]\subset]0,1[$, this means $\eta(s)=0 \Rightarrow Re(s)=\frac{1}{2}$ on the critical strip $Re(s)\in]0,1[$, which in turn means $\zeta(s)=0 \Rightarrow Re(s)=\frac{1}{2}$. Actually, I find that the "real part"'s asymptotic equivalent has to be, as $n \to \infty$: $$\frac{1}{2} + o\left(\frac{1}{\ln(n)^2}\right) + i2k\pi\left(\frac{\ln(n)-\gamma}{\ln(n)^2} + o\left(\frac{1}{\ln(n)^2}\right)\right), \space k \in \mathbb{Z}^*, \space \gamma=\lim_{n\to +\infty} \left(\sum_{k=1}^{n} \frac{1}{k} - \ln(n)\right)$$where $\gamma$ is the Euler-Mascheroni constant, And this "real part" eventually tends to $\frac{1}{2}$ as $n \to \infty$. NOTES: In my previous approach, I had expressed the problem as a system of quadratic equations, requiring all the $\Delta_{n-k}>0$, which required $r_n\sim -\frac{1}{n}$ which in turn ultimately required $\lim_{n\to\infty}a_n=\frac{1}{2}$, but after thorough verification, I figured out this would hold only if we imposed $r_n\neq -\frac{1}{k}, k\in \mathbb{N}^{*}$ and that there was no such constraint (or at least it's not obvious) This new approach proved to be more direct, without assuming such unverifiable constraints Compared to the previous version, I changed the way I managed the Taylor expansions to deduce the $\lambda + \frac{n^{1-2a_n}-1}{1-2a_n}\xrightarrow[n \to \infty]{} 0$ with $\lambda \in [0,1]$ condition; I had made a wrong simplification in previous versions. I also switched to Computer Modern Serif Font for a more Latex-like text. I'll keep making changes to add additional observations or correct some statements. Feedback would be appreciated at: [email protected] I am also looking forward to an arXiv endorsement, if anyone can help I sincerely thank them in advance.