# Show that the ratio of limits converges to the nearest Riemann zeta zero except when the ratio is a singularity

Let $$h(s,n)$$ be:

$$h(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-2}}{(n-2)!}\zeta (c)^{n-2} \sum _{k=1}^{n-1} \frac{(-1)^{k-1} \binom{n-2}{k-1}}{\zeta ((c-1) (k-1)+s)}$$

and let $$g(s,n)$$ be:

$$g(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-1}}{(n-1)!} \zeta (c)^{n-1} \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}$$

Question:

Does the ratio $$\rho(s) = i s+\lim\limits_{n \rightarrow \infty}\frac{h(i s,n)}{g(i s,n)}$$ converge to the nearest Riemann zeta zero?

For $$s=15$$ and $$n=12$$, we get: $$\rho(15) = 0.5 +14.1347 i$$

The first plot is the real part of $$\rho(s)$$, which begins at the trivial zero $$-2$$ and then tends to be close to $$1/2$$ except at singularities. The Gram points appear to be a subset of the singularities. The second plot is the imaginary part of $$\rho(s)$$, which has heights close to imaginary parts of Riemann zeta zeros.

(*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}]
(*end*)

(*start*)
(*Mathematica program for the first non trivial zeta zero*)
Clear[n, k, s, c, z, f, g];
n = 12;
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];
s = 15;
s*I + h[s*N[I]]/g[s*N[I]]
(*end*)


Clear[n, k, s, c];
n = 7;
s = N[14*I];
s - n*Limit[
1/Zeta[c]*
Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/
Sum[(-1)^(k - 1)*
Binomial[n, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n + 1}],
c -> 1]


For $$n=7$$ and $$s=14i$$:

$$0.5 + 14.1347i = s-n \left(\lim_{c\to 1} \, \frac{\sum _{k=1}^{n} \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}}{\zeta (c) \sum _{k=1}^{n+1} \frac{(-1)^{k-1} \binom{n}{k-1}}{\zeta ((c-1) (k-1)+s)}}\right)$$

The conjecture is that as $$n \rightarrow \infty$$ the limit above converges to the Riemann zeta zero nearest to $$s$$.

Set $$s=14.000000000000000000000000000000...i$$ with 1000 zeros after the decimal point. Set $$n=21$$ and set $$c = 1 + 1/10^{40}$$; With those parameters compute this formula:

$$s-\frac{n \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}}{\zeta (c) \sum _{k=1}^{n+1} \frac{(-1)^{k-1} \binom{n}{k-1}}{\zeta ((c-1) (k-1)+s)}}$$

What you will get is the 25 first decimal digits of the first Riemann zeta zero:

0.50000000000000000000000055508907479219367612957050478295942858083862
3727033228398609021142110650620136997773667771872221905026127340639625
41218507480832131294005829437
+
14.134725141734693790457251915896759601972505820234600660252328557362
5629956990194271674005286735176937891872097245657731536209606798029380
8035224527780328742481096881866 I

Of course ideally: $$n \rightarrow \infty$$ and $$c \rightarrow 1$$

(*Mathematica*)
(*start*)
Clear[n, k, s, c];
n = 21;
s = N[14*I, 1000];
c = 1 + 1/10^40;
s - n*(1/Zeta[c]*
Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n}]/
Sum[(-1)^(k - 1)*
Binomial[n, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n + 1}])
(*end*)


Here is my derivation.

A very short Mathematica program for computing the zeta zeros is:

Clear[x, t, nn];
nn = 12;
t = 15;
a = Series[1/Zeta[x + t*I], {x, 0, nn}];
t*I + N[Coefficient[a, x^(nn - 1)]/Coefficient[a, x^nn]]


which for $$t=15$$ gives $$0.5 + 14.1347i$$

Tom Copeland has recorded what he calls "Coefficients of list partition transform: reciprocal of an exponential generating function (e.g.f.)." in the OEIS here: https://oeis.org/A133314 together with several links to papers.

This is the table starting:

1
[-1]
[-1, 2]
[-1, 6, -6]
[-1, 8, 6, -36, 24]
[-1, 10, 20, -60, -90, 240, -120]
[-1, 12, 30, -90, 20, -360, 480, -90, 1080, -1800, 720]

These numbers above appear to be the same as the coefficients in the power series expansion of $$\frac{1}{f(x)} \tag{1}$$:
Which is given by the Mathematica command:

Series[1/f[x], {x, 0, 6}]

or as a table:
TableForm[CoefficientList[Series[1/f[x], {x, 0, 4}], x]]

$$\begin{array}{l} \frac{1}{f} \\ -\frac{f'}{f^2} \\ \frac{f'^2}{f^3}-\frac{f''}{2 f^2} \\ -\frac{f'^3}{f^4}+\frac{f' f''}{f^3}-\frac{f^{(3)}}{6 f^2} \\ \frac{24 f'^4-36 f f'^2 f''+6 f^2 f''^2+8 f^2 f' f^{(3)}-f^3 f^{(4)}}{24 f^5} \end{array}$$

This is of course is essentially the same as repeated derivatives of $$(1)$$ if one discards signs and multiply with factorials.
In Mathematica for the Riemann zeta function this would be:

Clear[s];
D[1/Zeta[s], {s, 0}]
D[1/Zeta[s], {s, 1}]
D[1/Zeta[s], {s, 2}]
D[1/Zeta[s], {s, 3}]
D[1/Zeta[s], {s, 4}]
D[1/Zeta[s], {s, 5}]


Mathematica knows that the first derivative of $$(1)$$ is: $$\frac{\partial \frac{1}{\zeta (s)}}{\partial s^1}=\frac{\zeta '(s)}{\zeta (s)^2}=\lim_{c\to 1} \, \left(\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}\right) \tag{2}$$

To get the second derivative we then recursively (as in repeated derivatives) nest the right hand side of $$(2)$$ into right hand side of $$(2)$$ to get:

$$\frac{\partial ^2\frac{1}{\zeta (s)}}{\partial s^2} = \frac{2 \zeta '(s)^2}{\zeta (s)^3}-\frac{\zeta ''(s)}{\zeta (s)^2}= \lim_{c\to 1} \, \left(\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}\right) \tag{3}$$

To get the third derivative we insert the right hand side of $$(3)$$ into the right hand side of $$(2)$$ to get:

$$\frac{\partial ^3\frac{1}{\zeta (s)}}{\partial s^3} = \frac{6 \zeta '(s)^3+\zeta ^{(3)}(s) \zeta (s)^2-6 \zeta (s) \zeta '(s) \zeta ''(s)}{\zeta (s)^4} = \lim_{c\to 1} \, \left(\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+c+s-1-1)}-\frac{\zeta (c)}{\zeta (c+c+c+s-1-1-1)}}}}}\right) \tag{4}$$

and so on...

This should be possible to show with some insertion of variables into the nested derivative limits. But I don't know how to do induction to prove it. And I have not yet inserted the variables, which probably should be inserted where there are free standing integers (in this case maybe the ones in the numerators).

In Mathematica this would be:

Expand[Limit[(Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]), c -> 1]]

Expand[Limit[(Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1), c -> 1]]

Expand[Limit[(Zeta[
c]/((Zeta[c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1),
c -> 1]]

Expand[Limit[(Zeta[
c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1 + c -
1]))^-1))^-1))^-1), c -> 1]]


Now we apply the Mathematica FullSimplify command to the expressions inside the limits:

FullSimplify[(Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1])]

FullSimplify[(Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1)]

FullSimplify[(Zeta[
c]/((Zeta[c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1)]

FullSimplify[(Zeta[
c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1 + c -
1]))^-1))^-1))^-1)]


This FullSimplify then gives us (to my surprise) for the right hand side of $$(2),(3)$$ and $$(4)$$:

$$\zeta (c) \left(\frac{1}{\zeta (s)}-\frac{1}{\zeta (c+s-1)}\right) \tag{from RHS of 2}$$ $$\zeta (c)^2 \left(\frac{1}{\zeta (s)}-\frac{2}{\zeta (c+s-1)}+\frac{1}{\zeta (2 c+s-2)}\right) \tag{from RHS of 3}$$ $$\zeta (c)^3 \left(\frac{1}{\zeta (s)}-\frac{3}{\zeta (c+s-1)}+\frac{3}{\zeta (2 c+s-2)}-\frac{1}{\zeta (3 c+s-3)}\right) \tag{from RHS of 4}$$ $$\zeta (c)^4 \left(\frac{1}{\zeta (s)}-\frac{4}{\zeta (c+s-1)}+\frac{6}{\zeta (2 c+s-2)}-\frac{4}{\zeta (3 c+s-3)}+\frac{1}{\zeta (4 c+s-4)}\right)$$

Apparently the numerators inside the parentheses are binomial coefficients with alternating signs and the denominators with the Riemann zeta function look like multiples of natural numbers. This leads us to the conjectured form:

$$g(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-1}}{(n-1)!} \zeta (c)^{n-1} \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}$$

when including signs and factorials. Because of the special limit for derivatives this formula works only for the Riemann zeta function. The Gamma function should give something similar.

n = 1;
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]


1/Zeta[s]

n = 2;
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]


-(Derivative[Zeta][s]/Zeta[s]^2)

n = 3;
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]


(2 Derivative[Zeta][s]^2 - Zeta[s] (Zeta^[Prime][Prime])[s])/(2 Zeta[s]^3)

n = 4;
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] • OEIS A133314 is precisely the entry for the reciprocal for an e.g.f. or Taylor series for an function f(x) analytic at the origin for which f(0)=1, as the title says. I retained the phrase "list partition transform" only for historical reasons. It's how I discovered it. – Tom Copeland Aug 7 '20 at 18:21
• Mathematica better give the same answer. – Tom Copeland Aug 7 '20 at 18:28
• I added a picture and some Mathematica code for the limits, for comparison. Seems to agree with the series expansion for 1/f(x), sofar. I plan to add this as either a comment or formula to A133314. – Mats Granvik Aug 7 '20 at 20:35
• May I suggest that you replace the codes with the corresponding mathematical statements in standard form and move the codes to github or other site with a link. (I used MathCad efficiently and effectively for years and don't relish learning yet another programming syntax/language/script). – Tom Copeland Aug 9 '20 at 21:33

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 the formula:

$$-\frac{\zeta '(s)}{\zeta (s)^2}=\lim_{c\to 1} \, \left(\frac{\zeta (c)}{\zeta (-(n-1)+(n-1)c+s)}-\frac{\zeta (c)}{\zeta (-n+nc+s)}\right)$$ for $$n=1,2,3,4,5,...$$

For $$n=1$$ the expression inside the limit is: $$A1=\left(\frac{\zeta (c)}{\zeta (-0+0c+s)}-\frac{\zeta (c)}{\zeta (-1+1c+s)}\right)$$ For $$n=2$$: $$A2=\left(\frac{\zeta (c)}{\zeta (-1+1c+s)}-\frac{\zeta (c)}{\zeta (-2+2c+s)}\right)$$ For $$n=3$$: $$A3=\left(\frac{\zeta (c)}{\zeta (-2+2c+s)}-\frac{\zeta (c)}{\zeta (-3+3c+s)}\right)$$ For $$n=4$$: $$A4=\left(\frac{\zeta (c)}{\zeta (-3+3c+s)}-\frac{\zeta (c)}{\zeta (-4+4c+s)}\right)$$

Then substitute to form the second derivative of $$(1)$$:

In $$A1$$ replace all $$\zeta(-1+c+s)$$ with $$\frac{1}{A2}$$ which results in:

$$B1=\frac{\zeta (c)}{\zeta (s)}-\zeta (c) \left(\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (2 c+s-2)}\right)$$

In $$B1$$ replace all $$\zeta(s)$$ with $$\frac{1}{A1}$$:

$$B2=\zeta (c) \left(\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}\right)-\zeta (c) \left(\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (2 c+s-2)}\right)$$

Here we FullSimplify $$B2$$ and get:

$$B3=\zeta (c)^2 \left(\frac{1}{\zeta (s)}-\frac{2}{\zeta (c+s-1)}+\frac{1}{\zeta (2 c+s-2)}\right)$$ which has binomial coefficients in the numerator. The limit:

$$\lim\limits_{c \rightarrow 1} B3 = \lim\limits_{c \rightarrow 1} \zeta (c)^2 \left(\frac{1}{\zeta (s)}-\frac{2}{\zeta (c+s-1)}+\frac{1}{\zeta (2 c+s-2)}\right)=\frac{2 \zeta '(s)^2-\zeta (s) \zeta ''(s)}{\zeta (s)^3} = \frac{\partial ^2\frac{1}{\zeta (s)}}{\partial s^2}$$

Mathematica puts it in a more readable form:

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];

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]


B1 = ReplaceAll[A1, Zeta[-1 + 1 c + s] -> 1/A2] means:
B1 equals the result of: "In A1 replace all Zeta[-1 + 1 c + s] with 1/A2"

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] $$\zeta (c)^5 \left(\frac{1}{\zeta (s)}-\frac{5}{\zeta (c+s-1)}+\frac{10}{\zeta (2 c+s-2)}-\frac{10}{\zeta (3 c+s-3)}+\frac{5}{\zeta (4 c+s-4)}-\frac{1}{\zeta (5 c+s-5)}\right)$$

This may be related

Hypergeometric-like Representation of the Zeta-Function of Riemann

where the Binomial expansion is used