A Possible Proof Of The Riemann Hypothesis

CORRECTIONS ONGOING: My assumption that the realness of all the $x_k$ required all the $\Delta_{n-k}\ge 0$ was obviously wrong; the correct constraint is "all the $\Delta_{n-k}\gt 0$ up to $k = n-O(1)$", and what really requires it are constraints on the $f_{n-k}$ and $g_{n-k}$ due to $0<x_k<1$ and $\left(\sum_{k=1}^{n} \frac{1}{k^{a_n}}\right)^2 \xrightarrow[n\to+\infty]{}+\infty$. I am currently correcting it. Updating as soon as possible. --- 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 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: $$ \sum_{k=1}^{n} \frac{1}{k^{2a_n}} + 2r_n\sum_{k=1}^{n-1} \sum_{j=k+1}^{n} \frac{1}{(kj)^{a_n}} \xrightarrow[n \to \infty]{} 0$$ $$\sum_{k=1}^{n} x_k^2 + 2r_n\sum_{k=1}^{n-1} \sum_{j=k+1}^{n} x_kx_j \xrightarrow[n \to \infty]{} 0, \space with \space x_1,\ldots,x_n \in \mathbb{R} \space and \space x_1=1,$$ --- CORRECTIONS ONGOING: My assumption that the realness of all the $x_k$ required all the $\Delta_{n-k}\ge 0$ was obviously wrong; the correct constraint is "all the $\Delta_{n-k}\gt 0$ up to $k = n-O(1)$", and what really requires it are constraints on the $f_{n-k}$ and $g_{n-k}$ due to $0<x_k<1$ and $\left(\sum_{k=1}^{n} \frac{1}{k^{a_n}}\right)^2 \xrightarrow[n\to+\infty]{}+\infty$. I am currently correcting it. Updating as soon as possible. --- The fact that all the $x_k$ are real requires all the corresponding $\Delta_k$ to be positive as well, this coupled with the condition that for all $n$ sufficiently large, $\forall k \in \{1,\ldots,n\}, x_k \gt 0$ then require: $$\Rightarrow r_n=-\frac{1}{n+\alpha} \space with \space \alpha \in ]-3,-1[$$ 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 two 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 $n \to \infty$: $$\frac{1}{2} + o\left(\frac{1}{\ln(n)^2}\right) + ik\pi\left(\frac{\ln(n)+4-\gamma}{\ln(n)^2} + o\left(\frac{1}{\ln(n)^2}\right)\right), \space k \in \mathbb{Z}^*, \space \gamma=\lim_{n\to +\infty} \left(\sum_{k=1}^{n} \frac{1}{k} - \ln(n)\right)$$where $\gamma$ is the Euler-Mascheroni constant,and that it therefore has an imaginary part that has a slower (so more prominent) rate of convergence $ik\pi\left(\frac{\ln(n)+4-\gamma}{\ln(n)^2}\right), \space k \in \mathbb{Z}^*, \space \gamma=\lim_{n\to +\infty} \left(\sum_{k=1}^{n} \frac{1}{k} - \ln(n)\right)$. And this imaginary rate of convergence in that "pseudo-real-part" vanishes as $n \to \infty$, yielding $\frac{1}{2}$ NOTES: I'll keep making changes to add additional observations or correct some statements. Feedback would be appreciated at: [email protected]