A Possible Proof Of The Riemann Hypothesis - A Partial Sums Approach

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 beginning 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 $\eta_n(s_n)$:

$$\sum_{k=1}^{n} \frac{1}{k^{2a_n}} + 2\sum_{1\le k \lt j \le n} \frac{(-1)^{k+j-2}\cos(b_n\ln(k/j))}{(kj)^{a_n}} = 0$$

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:

$$ \sum_{k=1}^{n} \frac{1}{k^{2a_n}} + 2r_n\sum_{1\le k \lt j \le n} \frac{1}{(kj)^{a_n}} = 0$$

$$ \Leftrightarrow \sum_{k=1}^{n} \frac{1}{k^{2a_n}} + 2r_n\left(\sum_{1\le k \lt j \le n} \frac{1}{(kj)^{a_n}}+\left(\sum_{k=1}^{n-1} \frac{1}{k^{a_n}}\right)\frac{1}{n^{a_n}}\right) = 0$$

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, in my isolated-sum-of-squares model, I find that the zero's asymptotic equivalent has to be, as $n \to \infty$: $$\frac{1}{2} + O\left(\frac{1}{\ln(n)^3}\right) + i\left(b_{\infty}+2k\pi\frac{\ln(n)-\gamma}{\ln(n)^2}+ O\left(\frac{1}{\ln(n)^3}\right)\right), \space k \in \mathbb{Z}^*, \space \gamma=\lim_{n\to +\infty} \left(\sum_{k=1}^{n} \frac{1}{k} - \ln(n)\right)$$
where $\gamma$ is the Euler-Mascheroni constant, and $b_{\infty}=\lim_{n\to\infty} b_n$.

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

In this version, I have taken into account the Euler-Maclaurin formula and realized my previous version wasn't correct, because there were hidden negligible terms that weren't negligible anymore after rescaling; I also had to add the third order term to the expansion for $n^{2-2a_n}-(n-1)^{2-2a_n}$ to have the more complete picture.

I'll keep making changes to add additional observations or correct some statements.

Feedback would be appreciated at: emrekaplan1978@gmail.com

I am also looking forward to an arXiv endorsement, if anyone can help I sincerely thank them in advance.