The limit in the question above can be found by simplifying ratios of consecutive repeated derivatives of $\frac{1}{\zeta(s)}$.

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My recent thoughts:

See: https://mathoverflow.net/a/439430/25104

Let:
$$\rho=\frac{1}{1-\frac{A}{B}}+\frac{1}{n}+s$$

Then according to Mathematica 8.0.1 the following:

$$\tag{1}$$
$$\text{Reduce}\left[\rho=\frac{1}{1-\frac{A}{B}}+\frac{1}{n}+s\land \frac{1}{1-\frac{A}{B}}+\frac{1}{n}+s=\left(-s+\frac{1}{1-\frac{B}{A}}-\frac{1}{n}\right)^*\land B\neq 0\land \Re(\rho)\geq 0\land \Re(\rho)\leq 1,\Re(\rho)\right]$$

reduces to the conditions:

$$\tag{2}$$
$A=\frac{B \cdot n \cdot \rho -B \cdot n \cdot s+B (-(n+1))}{n \cdot \rho-n \cdot s-1}$
$\land \left(\left(\Im(\rho)<\Im(s)\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\lor \left(\Im(\rho)=\Im(s)\land \left(\left(\Re(s)<\frac{n-2}{2 n}\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\lor \left(\Re(s)>\frac{n-2}{2 n}\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\right)\right)\lor \left(\Im(\rho)>\Im(s)\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\right)\land 0\leq \Re(\rho)\leq 1$

>So in other words there are choices of $A$, $B$ and the starting point $s$ where $\Re(\rho) = 1/2$, but the conditions above must be satisfied in order for the following equation:

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

to be true.  

In the case of the Riemann zeta function we would write:

$$A=\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s\right)} \tag{3}$$

$$B=\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s+\frac{1}{n}\right)} \tag{4}$$

The parentheses with the star:  
$$\left(\right)^*$$ is the complex conjugate.

How to show this in the case of the Riemann zeta function I don't know.

Demonstration in Mathematica 8.0.1:

    (* Mathematica start *)
    Clear[A, B, n, k, s];
    n = 20;
    s = 0;
    A = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n], {k, 1, n}];
    B = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n + 1/n], {k, 
        1, n}];
    "The first trivial zero with the zero of the conjugated limit shifted \
    by 1 and with opposite sign:"
    Conjugate[-s - 1/n + 1/(1 - B/A)];
    N[%, n]
     s + 1/n + 1/(1 - A/B);
    N[%, n]
    
    Clear[A, B, n, k, s];
    n = 20;
    s = (1/3 + 14*I);
    A = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n], {k, 1, n}];
    B = Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n + 1/n], {k, 
        1, n}];
    "The first non-trivial zero:"
    Conjugate[-s - 1/n + 1/(1 - B/A)];
    N[%, n]
     s + 1/n + 1/(1 - A/B);
    N[%, n]
    
    "The conditions to be proven for the case of the Riemann zeta function:"
    Clear[s, A, B, n, z, k];
    n = 19;
    Reduce[rho == s + 1/n + 1/(1 - A/B) && 
      s + 1/n + 1/(1 - A/B) == Conjugate[-s - 1/n + 1/(1 - B/A)] && 
      B != 0 && Re[rho] >= 0 && Re[rho] <= 1, Re[rho]]
    (*end*)