1:27 PM
(*start*)
Clear[s, c];
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];
(*start*)
(*Mathematica program for the plots*)
Clear[n, k, s, c, z, f, g];
n = 11;
ss = 40;
h[s_] = Limit[((-1)^(n - 2) Zeta[
c]^(n - 2) Sum[(-1)^(k - 1)*
Binomial[n - 2, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n - 1}]/(n - 2)!), c -> 1];
g[s_] = Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1];
Monitor[b = Table[s*I + h[s*N[I]]/g[s*N[I]], {s, 0, ss, 1/10}];, s*10]
ListLinePlot[Re[b], DataRange -> {0, ss}]
ListLinePlot[Im[b], DataRange -> {0, ss}]

1:40 PM
The plot of the real part starts at the trivial zero -2 and then stays close to 1/2. The staircase plot of the imaginary part has heights close to the imaginary parts of the Riemann zeta zeros.

3 hours later…
4:57 PM
@WolframMathematica ping!

5:12 PM
0

Here is a programmatically exact explanation of the derivation: First observe that the first derivative of: $$\frac{1}{\zeta(s)} \tag{1}$$ is: $$\frac{\partial \frac{1}{\zeta (s)}}{\partial s^1}=-\frac{\zeta '(s)}{\zeta (s)^2}$$ Mathematica knows that the first derivative can be computed through ...

2 hours later…