For what values of $c$ is $\sum _{k=1}^{\infty } (-1)^{k+1} x^{c \log (k)}=0$ when $x=\exp \left(-\frac{\rho _1}{c}\right)$?

Question

The alternating Dirichlet series, the Dirichlet eta function, can be written in the form: $\sum _{k=1}^{\infty } (-1)^{k+1} x^{c \log (k)}$

For what values of $c$ is $$\sum _{k=1}^{\infty } (-1)^{k+1} x^{c \log (k)}=0$$ when $x=\exp \left(-\frac{\rho _1}{c}\right)$, where $\rho_1$ is a Riemann zeta zero?

I conjecture that $c \geq \frac{\Im(\rho _1)}{\pi}$ is the right answer.

Setting $c=10$ and plotting convergents of Real and Imaginary parts:

Clear[n, k, x, c]
nn = 10^3;
c = 10;
x = N[Exp[-ZetaZero[1]/c], 100]
Plot[Re[Sum[(-1)^(k + 1)*x^(Log[k]*c), {k, 1, n}]], {n, 1, nn}]
Plot[Im[Sum[(-1)^(k + 1)*x^(Log[k]*c), {k, 1, n}]], {n, 1, nn}]
Sum[(-1)^(k + 1)*x^(Log[k]*c), {k, 1, Infinity}]

Answer 1

Note that with $\log$ the principal logarithm, one has $\log \exp z= z$ iff $-\pi < \Im z < \pi$, so if as usual $x^{c \log k}$ means that $\arg x \ne \pi$, so the principal $\log x$ is defined and $x^{c \log k}=\exp (c \log x \log k)$ then if $-\pi < \Im (-\rho_1/c) < \pi$ then your expression is just $\sum _{k=1}^{\infty } (-1)^{k+1} k^{-\rho_1}=0$ by the definition of $\rho_1$

Conversely, if $|\Im (\rho_1/c)| > \pi$ the series won't generally give zero since $-c \log x$ won't likely be a zero anymore but a translate by a non-zero multiple of $2\pi i c$, but one can write an equation for $c$ to get a zero occasionally (in other words one needs another zero $\rho_2$ on the same vertical line with $\rho_1$ and a $c$ st $\rho_1-\rho_2=2\pi i mc$ for some integral $m$ and also $\Im (-\rho_1/c)$ is in the same $m$ translate of the standard strip from $-\pi$ to $\pi$)

When $|\Im (\rho_1/c)| = \pi$ it depends on your defintions etc