The starting point for deriving the formula in the question is the following limits for the derivatives of the Riemann zeta function:
$$\zeta '(s)=\lim_{c\to 1} \, \zeta (s) \left(\zeta (c)-\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}\right)$$
$$\zeta ''(s)=\lim_{c\to 1} \, \zeta '(s) \left(\zeta (c)-\frac{\zeta (c) \zeta '(s)}{\zeta '(c+s-1)}\right)$$
$$\zeta ^{(3)}(s)=\lim_{c\to 1} \, \zeta ''(s) \left(\zeta (c)-\frac{\zeta (c) \zeta ''(s)}{\zeta ''(c+s-1)}\right)$$
$$\zeta ^{(4)}(s)=\lim_{c\to 1} \, \zeta ^{(3)}(s) \left(\zeta (c)-\frac{\zeta (c) \zeta ^{(3)}(s)}{\zeta ^{(3)}(c+s-1)}\right)$$
$$\zeta ^{(5)}(s)=\lim_{c\to 1} \, \zeta ^{(4)}(s) \left(\zeta (c)-\frac{\zeta (c) \zeta ^{(4)}(s)}{\zeta ^{(4)}(c+s-1)}\right)$$
$$\zeta ^{(6)}(s)=\lim_{c\to 1} \, \zeta ^{(5)}(s) \left(\zeta (c)-\frac{\zeta (c) \zeta ^{(5)}(s)}{\zeta ^{(5)}(c+s-1)}\right)$$
$$\vdots$$
$$\zeta ^{(n+1)}(s)=\lim_{c\to 1} \, \zeta ^{(n)}(s) \left(\zeta (c)-\frac{\zeta (c) \zeta ^{(n)}(s)}{\zeta ^{(n)}(c+s-1)}\right)$$
In mathematica this would be:
Clear[s, c]
Limit[((-Zeta[s]*Zeta[c]/Zeta[s + c - 1] + Zeta[c])*Zeta[s]), c -> 1]
Limit[((-Zeta'[s]*Zeta[c]/Zeta'[s + c - 1] + Zeta[c])*Zeta'[s]), c -> 1]
Limit[((-Zeta''[s]*Zeta[c]/Zeta''[s + c - 1] + Zeta[c])*Zeta''[s]), c -> 1]
Limit[((-Zeta'''[s]*Zeta[c]/Zeta'''[s + c - 1] + Zeta[c])*Zeta'''[s]), c -> 1]
Limit[((-Zeta''''[s]*Zeta[c]/Zeta''''[s + c - 1] + Zeta[c])*Zeta''''[s]), c -> 1]
Limit[((-Zeta'''''[s]*Zeta[c]/Zeta'''''[s + c - 1] + Zeta[c])*Zeta'''''[s]), c -> 1]
It is known that the ratios of consecutive derivatives a function in general converges to the nearest root (or singularity).
So:
$$\lim\limits_{n \rightarrow \infty} \frac{\zeta ^{(n+1)}(s)}{\zeta ^{(n)}(s)} + s = \text{nearest zero of the Riemann zeta function}$$
By succesively substituting the right hand side of the lower order derivatives of Riemann zeta into the right hand side of the immediate higher order derivative of Riemann zeta, with the help of the following Mathematica program;
Clear[s, c, A]
A0 = 1/Zeta[s];
Limit[Zeta[c] A0 - Zeta[c]/Zeta[-1 + c + s], c -> 1];
A1 = Zeta[c]/Zeta[-0 + 0 c + s] - Zeta[c]/Zeta[-1 + 1 c + s];
A2 = Zeta[c]/Zeta[-1 + 1 c + s] - Zeta[c]/Zeta[-2 + 2 c + s];
A3 = Zeta[c]/Zeta[-2 + 2 c + s] - Zeta[c]/Zeta[-3 + 3 c + s];
A4 = Zeta[c]/Zeta[-3 + 3 c + s] - Zeta[c]/Zeta[-4 + 4 c + s];
A5 = Zeta[c]/Zeta[-4 + 4 c + s] - Zeta[c]/Zeta[-5 + 5 c + s];
B1 = ReplaceAll[A1, Zeta[-1 + 1 c + s] -> 1/A2];
B2 = ReplaceAll[B1, Zeta[-0 + 0 c + s] -> 1/A1];
C1 = ReplaceAll[B2, Zeta[-2 + 2 c + s] -> 1/A3];
C2 = ReplaceAll[C1, Zeta[-1 + 1 c + s] -> 1/A2];
C3 = ReplaceAll[C2, Zeta[-0 + 0 c + s] -> 1/A1];
D1 = ReplaceAll[C3, Zeta[-3 + 3 c + s] -> 1/A4];
D2 = ReplaceAll[D1, Zeta[-2 + 2 c + s] -> 1/A3];
D3 = ReplaceAll[D2, Zeta[-1 + 1 c + s] -> 1/A2];
D4 = ReplaceAll[D3, Zeta[-0 + 0 c + s] -> 1/A1];
E1 = ReplaceAll[D4, Zeta[-4 + 4 c + s] -> 1/A5];
E2 = ReplaceAll[E1, Zeta[-3 + 3 c + s] -> 1/A4];
E3 = ReplaceAll[E2, Zeta[-2 + 2 c + s] -> 1/A3];
E4 = ReplaceAll[E3, Zeta[-1 + 1 c + s] -> 1/A2];
E5 = ReplaceAll[E4, Zeta[-0 + 0 c + s] -> 1/A1];
FullSimplify[A0]
FullSimplify[A1]
FullSimplify[B2]
FullSimplify[C3]
FullSimplify[D4]
FullSimplify[E5]
one gets:
FullSimplify[A0]
$$\frac{1}{\zeta (s)}$$
FullSimplify[A1]
$$\zeta (c) \left(\frac{1}{\zeta (s)}-\frac{1}{\zeta (c+s-1)}\right)$$
FullSimplify[A2]
$$\zeta (c)^2 \left(\frac{1}{\zeta (s)}-\frac{2}{\zeta (c+s-1)}+\frac{1}{\zeta (2 c+s-2)}\right)$$
FullSimplify[A3]
$$\zeta (c)^3 \left(\frac{1}{\zeta (s)}-\frac{3}{\zeta (c+s-1)}+\frac{3}{\zeta (2 c+s-2)}-\frac{1}{\zeta (3 c+s-3)}\right)$$
FullSimplify[A4]
$$\zeta (c)^4 \left(\frac{1}{\zeta (s)}-\frac{4}{\zeta (c+s-1)}+\frac{6}{\zeta (2 c+s-2)}-\frac{4}{\zeta (3 c+s-3)}+\frac{1}{\zeta (4 c+s-4)}\right)$$
FullSimplify[A5]
$$\zeta (c)^5 \left(\frac{1}{\zeta (s)}-\frac{5}{\zeta (c+s-1)}+\frac{10}{\zeta (2 c+s-2)}-\frac{10}{\zeta (3 c+s-3)}+\frac{5}{\zeta (4 c+s-4)}-\frac{1}{\zeta (5 c+s-5)}\right)$$
That looks like the binomial coefficients, and should be because of the binary(?) substition.
Imitating the last formula:
Clear[n, k, c, h, s]
n = 6;
Table[(-1)^(k + 1)*
Binomial[n - 1, k - 1]/Zeta[-(k - 1) + (k - 1) c + s], {k, 1, n}]
So:
$$\frac{\zeta (c)^{n-1} \sum _{k=1}^n \frac{(-1)^{k+1} \binom{n-1}{k-1}}{\zeta (c (k-1)-(k-1)+s)}}{\zeta (c)^{n+1-1} \sum _{k=1}^{n+1} \frac{(-1)^{k+1} \binom{n+1-1}{k-1}}{\zeta (c (k-1)-(k-1)+s)}} \tag{1}$$
One can strike out part of the binomials to get the $\frac{A}{B}$ formula in the question.
How ever, in order to not lose information, one sets $c \rightarrow 1+\frac{1}{h}$
Because both $n \rightarrow \infty$ and $h \rightarrow \infty$ and we want to have a formula with as few variables as possible, we simply replace $h$ with $n$ (that is, $h=n$ from now on, because we say so). So:
$$c=1+\frac{1}{n}$$
close to $c=1$: $\zeta (c) \approx \frac{1}{c-1}$
introducing these changes, except striking out binomials, into $(1)$ we get:
$$\frac{\sum _{k=1}^n \frac{(-1)^{k+1} \binom{n-1}{k-1}}{\zeta \left(\frac{k-1}{n}+s\right)}}{\zeta \left(1+\frac{1}{n}\right) \left(\sum _{k=1}^{n+1} \frac{(-1)^{k+1} \binom{n}{k-1}}{\zeta \left(\frac{k-1}{n}+s\right)}\right)}$$
Striking out binomials we get:
$$\Re\lim_{n \rightarrow \infty}
\left(
\left[
1-
\left(
\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta(\tfrac{k}{n}+s)}
\Bigg/
\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta(\tfrac{k}{n}+s+\tfrac{1}{n})}
\right)
\right]^{-1}
\underset{\text{conjectured guesses}}{\underbrace{+\frac1n + s}}
\right) = \frac{1}{2}.$$
with conjectured terms $$+\frac1n + s$$ that are guesses to make it work.
We now have the formula in the form:
$$\rho=\frac{1}{1-\frac{A}{B}}+\frac{1}{n}+s$$
Please go to the previous comment as an answer: https://mathoverflow.net/a/377346/25104
