The Dirichlet eta function is:
$$\eta (s) = \zeta (s) (1-2^{1-s})$$ $$\eta (s) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^{s}}$$
Notice that:
$$\log(k)=\lim\limits_{n \rightarrow \infty} \frac{\text{Round}[n \log (k)]}{n}$$
$$k^{s} = (e^{\log(k)})^{s}$$
$$k^{s}=\lim\limits_{n \rightarrow \infty} {\left(e^{\frac{\text{Round}[n \log (k)]}{n}}\right)^{s}}$$ so that the Dirichlet eta function becomes:
$$\eta (s) = \lim\limits_{n \rightarrow \infty} \sum_{k=1}^{n} \frac{(-1)^{k+1}}{\left(e^{\frac{\text{Round}[n \log (k)]}{n}}\right)^{s}}$$
or alternatively: $$\eta (s) = \lim\limits_{n \rightarrow \infty} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{\left(e^{\frac{\text{Round}[n \log (k)]}{n}}\right)^{s}}$$
Solving $\text{the Right Hand Side}=0$ for $s$:
$$n \left(2 \pi i C-\log \left(\text{Root}\left[\sum _{k=1}^n (-1)^{k+1} s^{\text{Round}[n \log (k)]}\right]\right)\right)$$
Since we in the Riemann hypothesis are interested only in the real part, we can discard the integer multiples $C$ of $2 \pi i$:
$$-n \log \text{Root}\left[\sum _{k=1}^n (-1)^{k+1} s^{\text{Round}[n \log (k)]}\right]$$
Without the logarithm we then have the $m-th$ root as:
$$z_{m} = \text{Root}\left[\sum _{k=1}^n (-1)^{k+1} s^{\text{Round}[n \log (k)]}\right]$$
Substituting $s \rightarrow x$ and considering only the remaining polynomial we then have:
$$P(x) = \sum_{k=1}^{n} (-1)^{k+1} x^{\text{Round}[n \log (k)]}$$
or alternatively:
$$P(x) = \sum_{k=1}^{\infty} (-1)^{k+1} x^{\text{Round}[n \log (k)]}$$
If we compute $-n\log(z_m)$ where $m=\text{Round}\left[\log(n)n\frac{3}{4}\right]..\text{Round}\left[\log(n)n\right]$, of the polynomial:
$$P(x) = \sum _{k=1}^n (-1)^{k+1} x^{\text{Round}[n \log (k)]} \tag{$*$}$$
we see that the roots cluster in the complex plane around the vertical line with real parts: $$\Re(-n\log(z_j)) \approx \frac{1}{2}$$ and $$\Re(-n\log(z_t)) \approx 1$$ (for some $j \in m$ and $t \in m$) as can be seen in this plot for the polynomial for $n=500$:
(*Mathematica 8 start*)
nn = 100;
n = nn;(*Increase n=500 for better precision*)
$MaxRootDegree = Round[Log[n]*n] + 1;
Monitor[z =
Table[(N[
Root[Sum[(-1)^(k + 1)*#1^Round[(Log[k]*n)], {k, 1, n}] &,
m]]), {m, Round[Log[n]*n*3/4], Round[Log[n]*n]}];, n]
ListPlot[Chop[
Table[{Re[-nn*Log[z[[n]]]], Im[-4/Pi*Log[z[[n]]]]}, {n, 1,
Length[z]}]]]
ListPlot[Chop[Table[{Re[z[[n]]], Im[z[[n]]]}, {n, 1, Length[z]}]],
AspectRatio -> 4]
(*end*)
The imaginary part of the logarithm of the zeros is multiplied with minus four divided by $\pi$ as: $\Im\left(-\frac{4}{\pi}\log(z_i)\right)$ so that the plot is normalized to height one.
The polynomial roots without taken the logarithm of, form a quarter segment of a circle:
If the we instead plot the logarithm of all the polynomial zeros $-n\log(z_m)$ for $n=100$ and $m=1..\text{Round}\left[\log(n)n\right]$ we get a more spread out plot:
and a complete but rough circle (when plotting the roots $z_m$ of the polynomial $P(x)$ in $(*)$ directly, and not taking logarithms of them):
Now since there is the Theorem 1.1. for polynomials of even degree in the pdf notes by Keith Conrad at: https://kconrad.math.uconn.edu/blurbs/galoistheory/numbersoncircle.pdf
Theorem 1.1. Let $f(z) \in \mathbb{Q}\left[z\right]$ be irreducible with degree $n > 1$. If $f(z)$ has a root on the unit circle then $n$ is even and $z^n f(1/z) = f(z)$,
I will ask if such theorems can be used to prove that one or several roots of the polynomial in $(*)$:
$$P(x) = \sum _{k=1}^n (-1)^{k+1} x^{\text{Round}[n \log (k)]}$$ lie on the critical circle when $n \rightarrow \infty$, instead of the critical line as in the Riemann hypothesis?
The radius of the critical circle should yet be computed. This builds upon the answer by Professor Jeffrey Stopple at MathOverflow: https://mathoverflow.net/a/398389/25104
Any rational approximation $p/q$ of $\log(k)$ will do, but this one above I figured is the fastest. I have also considered using truncated Dirichlet series expansions of the fundamental theorem of arithmetic: $\sum_{d|k} \Lambda(d) = \log(k) $, where $\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$. In that way one gets several separate lines in the root plot in the complex plane, instead of a root cloud clustering around $\Re(-n\log(s)) = 1/2$.
But why does SumConvergence[(-1)^(k + 1)*x^Log[k], k] give
1 + Log[Abs[x]] < 0?