The limit in the question above can be found by simplifying ratios of consecutive repeated derivatives of $\frac{1}{\zeta(s)}$.
My recent thoughts:
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:
$$\tag{2}$$ $A=\frac{B \cdot n \cdot \rho -B \cdot n \cdot s+B (-(n+1))}{n \cdot \rho-n \cdot s-1}$ $\land \left(\left(\Im(\rho)<\Im(s)\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\lor \left(\Im(\rho)=\Im(s)\land \left(\left(\Re(s)<\frac{n-2}{2 n}\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\lor \left(\Re(s)>\frac{n-2}{2 n}\land \Re(\rho)=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\right)\right)\lor \left(\Im(\rho)>\Im(s)\land \Re(\text{rho})=\frac{1}{2}\land (\Re(B)<0\lor (\Re(B)=0\land (\Im(B)<0\lor \Im(B)>0))\lor \Re(B)>0)\right)\right)\land 0\leq \Re(\rho)\leq 1$
So in other words there are choices of $A$, $B$ and the seed 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[z == 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[z] >= 0 && Re[z] <= 1, Re[z]]
(*end*)