Question concerning an assertion regarding the modulus of the Riemann Zeta function (follow up)

Question

Update December 2023 -- Some additional cleanup, however the general argument remains the same. I have yet to see these telescoping/collapsing equations in the literature. Do reach out via email if you would like to collaborate or you know of someone who might be able to assist.

Posting this question here as the community has addressed the Riemann Zeta function with questions such Simpler zeta zeros and Zeta function zeros and analytic continuation.

This is my second post, see Question concerning an assertion regarding the modulus of the Riemann Zeta function.

Define the following Dirichlet series, i.e. partial sum of the Riemann Zeta function as \begin{equation} \zeta_{n}(s)=\zeta_{n}(\sigma+it)=\sum_{k=1}^{n}\frac{1}{k^{\sigma+it}} \end{equation}

I take it known that in the critical strip that using the Euler-Maclaurin formula, Hardy and Littlewood showed that for a large $n\in\mathbb{N}$: \begin{align} \zeta(s)=\zeta_{n}(s)-\frac{n^{1-s}}{1-s}+O(n^{-\sigma}) \end{align}

The expression $\frac{n^{1-s}}{1-s}$ describes the specific logarithmic spiral that the polylines of $\zeta_{n}(s)$ approximate, but always circumnavigates the origin. It follows that if and only if $\sigma+it$ is a non-trivial zeta zero, the polylines formed from $\zeta_{n}(\sigma+it)$ spiral $0+0i$, this has been noted elsewhere, see the paper "A Geometric Perspective on the Riemann Zeta Function’s Partial Sums" by Carl Erickson (https://dl.icdst.org/pdfs/files3/46b9a82eac445746e69d8e6d76c17749.pdf).

The Riemann Zeta function can be expressed by isolating the real and complex parts of the function as \begin{equation} \zeta(\sigma+it)=\sum_{n=1}^{\infty}\frac{\cos(t\ln(n))}{n^\sigma}-i\sum_{n=1}^{\infty}\frac{\sin(t\ln(n))}{n^\sigma} \end{equation}

Take $\sigma+it$ in the critical strip, then the modulus (distance from the origin and $\zeta_{n}(\sigma+it)$) can be computed as:

\begin{align} |\zeta_{n}(\sigma+it)|&=\sqrt{\left(\sum_{1}^{n}\frac{\cos(t\ln(n))}{n^\sigma}-0\right)^2+\left(\sum_{1}^{n}\frac{\sin(t\ln(n))}{n^\sigma}-0\right)^2}\\ &=\sqrt{\sum_{j=1}^{n}\sum_{k=1}^{n}\frac{\cos(t\ln(j))\cos(t\ln(k))}{(jk)^\sigma}+\sum_{j=1}^{n}\sum_{k=1}^{n}\frac{\sin(t\ln(j))\sin(t\ln(k))}{(jk)^\sigma}}\\ \normalsize &=\sqrt{\sum_{j=1}^{n}\sum_{k=1}^{n}\frac{\cos(t(\ln(j)-\ln(k)))}{(jk)^\sigma}}\\ &=\sqrt{\sum_{j=1}^{n}\sum_{k=1}^{n}\frac{\cos(t\ln(\frac{j}{k}))}{(jk)^\sigma}} \end{align}

A similar (alternating version) of this last equation is noted in "Asymptotic formulas for harmonic series in terms of a non-trivial zero on the critical line" by Arhur Kawalec (https://arxiv.org/pdf/1911.06115.pdf).

As a side note, $|\zeta_{n}(s)|$ is an interesting equation, it describes something unique to the Zeta function by combining the natural logarithm, trigonometry, and the set of rationals (i.e. the collection of $\frac{j}{k}$). As noted on my previous question, from observation/computation this function $|\zeta_{n}(s)|$ is monotonic if and only if $s$ is a non-trivial zeta zero-- albeit I don't have proof. This is counter-intuitive, and a bit wild given the cosine function. Do reach out if you have ideas here.

Next, define the following function $g_{n}(s)$ which identifies the summands, before the square root: \begin{equation} g_{n}(s)=|\zeta_{n+1}|^2-|\zeta_{n}|^2=2\left(\sum_{j=1}^{n}\frac{\cos(t\ln(\frac{j}{n}))}{(jn)^\sigma}\right)+\frac{1}{n^{2\sigma}} \end{equation}

Define the following function $f_n(s)$ which is the change in modulus from the $n^{\text{th}}$ to the $n+1^{\text{th}}$ term: \begin{equation} f_{n}(s)=|\zeta_{n+1}(s)|-|\zeta_{n}(s)|=\sqrt{\sum_{j=1}^{n}\sum_{k=1}^{n}\frac{\cos(t\ln(\frac{j}{k}))}{(jk)^\sigma}+g_{n}(s)}-\sqrt{\sum_{j=1}^{n}\sum_{k=1}^{n}\frac{\cos(t\ln(\frac{j}{k}))}{(jk)^\sigma}} \end{equation}

Lemma 1 Suppose $s$ is some some non-trivial zeta zero with $\sigma>\frac{1}{2}$, then $n$ can be chosen large enough such that $ \lim_{n\rightarrow\infty}\sum_{k=1}^{n}g_{k}(s)=\lim_{n\rightarrow\infty}\sum_{k=1}^{mn}g_{k}(s)$, $m\in\mathbb{N^{+}}$.

(Geometrically, with $\sigma>\frac{1}{2}$ in the critical, the rate that the modulus grows, decreases towards $0$ as $n$ increases. We use this fact, to state that for $n$ large enough, some $m$ can be chosen such that the modulus at $n$ steps and $mn$ steps is within any chosen $\varepsilon>0$.)

Proof Assuming that $n$ is large enough to be circumnavigating the final value of the $\zeta(s)$, consider a significantly larger number of polylines, i.e $\zeta_{mn}$, $m\in\mathbb{N}^+$. By choosing $mn$ instead of say $2n$, the Hardy-Littlewood spiral comes into play to distinguish between non-trivial zeta zeros and other points in the critical strip.

Consider first if $s$ is a non-trivial zero, then $\zeta(s)=0$ yields $0=\zeta_{mn}(s)-\frac{(mn)^{1-s}}{1-s}+O((mn)^{-\sigma})$, which shows that as $n\rightarrow\infty$, the rate at which the modulus of $\zeta_{mn}$ spirals outward from $0$ slows (save $O((mn)^{-\sigma})$ which will also approach zero as $n\rightarrow\infty$). As $n\rightarrow\infty$, the spiral expands outwards slower and slower from the origin-- for any $\varepsilon>0$, $n$ can be chosen large enough such that on the next circumnavigation around the origin (some $mn$), the magnitude of the polylines can be within the chosen $\varepsilon$.

Next consider some $s'$ as some complex number, with $\frac{1}{2}<\sigma<1$, that is not a non-trivial zeta zero, then the equation $\zeta(s)=\zeta_{mn}(s)-\frac{(mn)^{1-s}}{1-s}+O((mn)^{-\sigma})$ does not simplify in the same manner. From $n$ to $mn$, the value of $\zeta_{mn}(s')$ may be as much as $2|\zeta(s')|+O((mn)^{-\sigma})$ away from $\zeta_{n}$, given that $\zeta(s')\neq0$, the modulus will oscillate, not converging in the same manner.

Lemma 2 The $\zeta_{mn}$ summation, $\sum^{mn}_{j=1}\frac{\cos(t\ln(\frac{j}{mn}))}{(j mn)^\sigma}$ can be approximated within any $\varepsilon>0$ by $\sum^{n}_{k=1}m\frac{\cos(t\ln(\frac{mk}{mn}))}{(mk\cdot mn)^\sigma}$ as $n\rightarrow\infty$.

This assertion can be formalized using Riemann integration.

Finally, the curious bit, and the entire purpose of all of the above, is the following from the sums from this $g_n(s)$. I start by simply considering $s$ as some non-trivial zeta zero, $\frac{1}{2}<\sigma<1$. I then add polylines out from $\zeta_{n}$ to $\zeta_{mn}$, using Lemma 1 and Lemma 2 to telescope out and in from $n$ summands to $mn$, and back to $n$ summands, followed by some algebra:

\begin{align} \lim_{n\rightarrow\infty}\sum^{n}_{j=1}\frac{\cos(t\ln(\frac{j}{n}))}{(j n)^\sigma}\\ &=\frac{\cos(t\ln(\frac{1}{n}))}{(1\cdot n)^\sigma}+\frac{\cos(t\ln(\frac{2}{n}))}{(2\cdot n)^\sigma}+\cdots+\frac{\cos(t\ln(\frac{n}{n}))}{(n\cdot n)^\sigma}\\ &=\frac{\cos(t\ln(\frac{1}{mn}))}{(1\cdot mn)^\sigma}+\frac{\cos(t\ln(\frac{2}{mn}))}{(2\cdot mn)^\sigma}+\cdots+\frac{\cos(t\ln(\frac{m}{mn}))}{(m\cdot mn)^\sigma}+\cdots+\frac{\cos(t\ln(\frac{mn}{mn}))}{(mn\cdot mn)^\sigma}\\ &=m\left(\frac{\cos(t\ln(\frac{m}{mn}))}{(m\cdot mn)^\sigma}\right)+m\left(\frac{\cos(t\ln(\frac{m2}{mn}))}{(m2\cdot mn)^\sigma}\right)+\cdots+m\left(\frac{\cos(t\ln(\frac{mn}{mn}))}{(mn\cdot mn)^\sigma}\right)\\ &=\lim_{n\rightarrow\infty}\frac{m}{m^{2\sigma}}\sum^{n}_{j=1}\frac{\cos(t\ln(\frac{j}{n}))}{(j n)^\sigma} \end{align}

Taken together, \begin{align} \lim_{n\rightarrow\infty}\sum^{n}_{j=1}\frac{\cos(t\ln(\frac{j}{n}))}{(j n)^\sigma}&=\frac{m}{m^{2\sigma}}\lim_{n\rightarrow\infty}\sum^{n}_{j=1}\frac{\cos(t\ln(\frac{j}{n}))}{(j n)^\sigma}\\ 1&=\frac{m}{m^{2\sigma}}\\ \sigma&=\frac{1}{2} \end{align}

Given that there is symmetry around the line $\sigma=\frac{1}{2}$ (see Proof of Symmetry of zeros in critiical strip), it follows from the above that if $\zeta(s)=0$, then the real part of $s$ must be within any $\varepsilon>0$ of $\frac{1}{2}$.

My question: is this not a nice geometric illustration that the RH is true?

Answer 1

Let me put some comments: first if $\sigma >1/2$ we have $\lim_{n \to \infty}g_n(s)=0$ since $|1-s||\zeta_n(s)|=n^{1-\sigma}+O(n^{-\sigma})$ so $|\zeta_n(s)|-|\zeta_{n+1}(s)|=O(n^{-\sigma})$ while $|\zeta_n(s)|+|\zeta_{n+1}(s)|=O(n^{1-\sigma})$ so their product is $O(n^{1-2\sigma})=o(1)$ hence even if the rest is correct you just get $0=m^{1-2\sigma} \times 0$ which tells you nothing about $\sigma$ in this case.

If $\sigma=1/2$ it is unclear to me why the limit of $g_n(s)$ exists since in that case $|\zeta_n(s)|-|\zeta_{n+1}(s)|=O(1/\sqrt n)$ and $|\zeta_n(s)|+|\zeta_{n+1}(s)|=cn^{1/2}+O(1/\sqrt n)$ but we have no control of the error term in the original equation and $|\zeta_n(s)|-|\zeta_{n+1}(s)|$ is of the same size with it, hence we just get that $g_n(s)$ is bounded, but convergence requires proof.

For $\sigma <1/2$ similarly one shows that $\limsup g_n(s)=\infty$ (I suspect that $g_n(s) \to \infty$ can be proven with a careful analysis of the error term in the equation above) so again you get the less than illuminating $\infty =m^{1-2\sigma} \times \infty$