Solutions to $\frac{1}{1-\frac{A}{B}}+s=\left(\frac{1}{1-\frac{B}{A}}-s\right)^*$ over the Complexes.

Question

Here I describe how I arrived at this equality, or its starting point:

$$\lim_{n\to \infty } \, \left(\frac{1}{1-\frac{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s-\frac{1}{n}\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s\right)}}}+s\right)=\lim_{n\to \infty } \, \left(\frac{1}{1-\frac{\sum _{k=1}^n \frac{(-1)^{k-1} \left( \begin{array}{c} n-1 \\ k-1 \end{array} \right)}{\zeta \left(\frac{k}{n}+s\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \left( \begin{array}{c} n-1 \\ k-1 \end{array} \right)}{\zeta \left(\frac{k}{n}+s-\frac{1}{n}\right)}}}-s\right){}^* \tag{1}$$

which for $s=14i$ numerically evaluates to:

$$0.5000000000... + i14.134725141... = (0.5000000000... - i14.134725141... )^* \tag{2}$$

which is the first Riemann zeta zero.

Is there any way to prove $(1)$?

The equality above has the form:

$$\lim_{n\to \infty } \, \left(\frac{1}{1-\frac{A(n,s)}{B(n,s)}}+s\right)=\lim_{n\to \infty } \, \left(\frac{1}{1-\frac{B(n,s)}{A(n,s)}}-s\right)^*$$

where: $$A(\text{n},\text{s})\text{=}\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}-\frac{1}{n}+s\right)}$$ $$B(\text{n},\text{s})\text{=}\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}+s\right)}$$ where: $$f(\text{x})\text{=}\zeta (x)$$

Therefore I entered this in Mathematica:

Clear[f, A, B, x, n, k, s, rho];
f[x_] := Zeta[x];
A[n_, s_] := 
 Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/f[s + k/n - 1/n], {k, 1, n}]
B[n_, s_] := 
 Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/f[s + k/n], {k, 1, n}]
n = 30;
s = 14*I;
rho = s + 1/(1 - A[n, s]/B[n, s]);
N[%, n]
rho = -s + 1/(1 - B[n, s]/A[n, s]);
N[%, n]

giving:

0.5000000000000000000000000000 + 14.1347251417346937904572519836 I

0.5000000000000000000000000000 - 14.1347251417346937904572519836 I

as well as this:

Clear[f, A, B, x, n, k, s, rho];
Reduce[rho == s + 1/(1 - A/B) && 
  s + 1/(1 - A/B) == Conjugate[-s + 1/(1 - B/A)], Re[rho], Complexes]

which is the simplified equality:

$$\left(\frac{1}{1-\frac{A}{B}}+s\right)=\left(\frac{1}{1-\frac{B}{A}}-s\right)^*$$

treating $A(n,s)$ as simply the quantity $A$, and $B(n,s)$ as the quantity $B$, giving:

(B == 0 && ((Re[A] < 0 && 
       Re[s] == (Im[A]^2 + Re[A]^2)/(2 Im[A]^2 + 2 Re[A]^2) && 
       rho == I Im[s] + Re[s]) || (Re[A] == 
        0 && ((Im[A] < 0 && Re[s] == 1/2 && 
           rho == 1/2 + I Im[s]) || (Im[A] > 0 && Re[s] == 1/2 && 
           rho == 1/2 + I Im[s]))) || (Re[A] > 0 && 
       Re[s] == (Im[A]^2 + Re[A]^2)/(2 Im[A]^2 + 2 Re[A]^2) && 
       rho == I Im[s] + Re[s]))) || (((Im[rho] < Im[s] && 
       Re[rho] == 1/2
        && (Re[B] < 0 || (Re[B] == 0 && (Im[B] < 0 || Im[B] > 0)) ||
          Re[B] > 0)) || (Im[rho] == 
        Im[s] && ((Re[s] < 1/2 && 
           Re[rho] == 1/2
            && (Re[B] < 
              0 || (Re[B] == 0 && (Im[B] < 0 || Im[B] > 0)) || 
             Re[B] > 0)) || (Re[s] > 1/2 && 
           Re[rho] == 1/2
            && (Re[B] < 
              0 || (Re[B] == 0 && (Im[B] < 0 || Im[B] > 0)) || 
             Re[B] > 0)))) || (Im[rho] > Im[s] && 
       Re[rho] == 1/2
        && (Re[B] < 0 || (Re[B] == 0 && (Im[B] < 0 || Im[B] > 0)) ||
          Re[B] > 0))) && A == (-B + B rho - B s)/(rho - s))

Where && means AND and || means OR.

I would have posted it as latex but it stretches over the border to the right and does not fit the page because I don't know how insert line break for such a long expression.

$()^*$ is the complex conjugate, and Mathematica gives the solutions over the Complexes.

What I notice is that there are scenarios for the zeta zeros having real part equal to $\frac{1}{2}$, according to the Mathematica output, see Re[rho] == 1/2, but it all depends on the equality of limits $(1)$ above being true. So again, is there any way to prove $(1)$?