Posting this question here as the community has addressed the Riemann Zeta function with questions such as, i.e. with Simpler zeta zeros and Zeta function zeros and analytic continuation.
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 the origin, 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) which has been referenced by members of these boards.
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 $\theta+it$ as some non-trivial zeta zero, then the modulus (distance from the origin, $\zeta(\theta+it)$, and $\zeta_{n}(\theta+it)$) 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}
An similar/ alternating version of this last equation is noted in "Asymptotic formulas for harmonic series in terms of anon-trivial zero on the critical line" by Arhur Kawalec (https://arxiv.org/pdf/1911.06115.pdf).
As a side note before my question, this is a beautiful 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 $\frac{i}{j}$).
My first question, have I made any mistakes yet?
Take $\theta+it$ as some non-trivial zeta zero, for any $\varepsilon>0$, an $n\in\mathbb{N}$ can be chosen large enough such that the logarithmic spiral above has the same modulus (within $\varepsilon$) of $|\zeta_{n}(\sigma+it)|$, for any and all integers greater than $n$. [Given the curvature of said logarithmic spiral in the critical strip,] I believe that I can now claim that for any $\varepsilon>0$, $||\zeta_{n+1}(\sigma+it)|-|\zeta_{n}(\sigma+it)||<\varepsilon$, for any integers greater than $n$. Note that this claim cannot be made if $\theta+it$ is not a zeta zero as the partial sums of $\zeta_{n}$ won't spiral the origin, see the two images here to help visualize geometrically what is happening:
log spiral approximating (and not approximating) the partial sums of the zeta function
My second question: is this previous assertion correct?
Geometrically (see image), it seems to me that a distinction can be made though, while both are within $O(n^{-\sigma})$ when $s$ is a zeta zero, and when $s$ is not a zero. Geometrically (& and even through computation), the series $|\zeta_{n}(s)|, |\zeta_{n+1}(s)|, |\zeta_{n+2}(s)| $ is monotonic if and only if $s$ is a zero. Put another way, while both are less than $O(n^{-\sigma})$, if $s$ is a zero then the modulus would be significantly smaller than just $O(n^{-\sigma})$ (after $n$ is large enough).
– jwren Nov 09 '23 at 17:31