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Mats Granvik
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Let: $$f(x)=\zeta (x)$$ $$A(n,s)=\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}-\frac{1}{n}+s\right)}$$ $$B(n,s)=\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}+s\right)}$$ $$X(n,s)=\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}-\frac{1}{n}-s\right)}$$ $$Y(n,s)=\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}-s\right)}$$ $$a=\frac{1}{1-\frac{A(n,s)}{B(n,s)}}+s$$ $$b=\frac{1}{1-\frac{B(n,s)}{A(n,s)}}-s$$ $$c=\frac{1}{1-\frac{X(n,s)}{Y(n,s)}}-s$$ Notice that: $$X(n,s)=A(n,-s)$$
and
$$Y(n,s)=B(n,-s)$$

For $s=1/3+14i$, show: $$\lim_{n\to \infty } \, ((a+b)(1-(b-c)))=1$$ Leaving out the limit symbol and substituting $a,b,c$: $$\left(\frac{1}{1-\frac{A}{B}}+\frac{1}{1-\frac{B}{A}}\right) \left(-\frac{1}{1-\frac{B}{A}}+\frac{1}{1-\frac{X}{Y}}+1\right)=1$$ which is equal to: $$-\frac{A Y+B X-2 B Y}{(A-B) (X-Y)}=1$$ multiplying with the denominator: $$-A Y-B X+2 B Y=(A-B) (X-Y)$$ subtracting with the right hand side: $$-A Y-B X+2 B Y -(A-B) (X-Y)=0$$ factoring: $$B Y-A X=0$$ which is: $$A X=B Y$$ Including the limit symbol again and substituting $A,B,X,Y$: $$\lim_{n\to \infty } \, \left(\left(\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}-\frac{1}{n}+s\right)}\right) \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}-\frac{1}{n}-s\right)}=\left(\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}+s\right)}\right) \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}-s\right)}\right)$$ For $n=30$ and $s=1/3+14i$ we get numerically:
for the left hand side:
$$55226.0411027488837442269063281-14296.8517199926101555805382701 i$$ and for the right hand side:
$$55226.0411027488837442269063281-14296.8517199926101555805382701 i$$ which appear close to each other.


while for example if we set the function to $f(x)=\zeta \left(x,\frac{1}{3}\right)$ we get numerically,
for the left hand side: $$-\text{5.7804095358568700751853287633386056719879460800172587438796357645$\grave{ }$30.044170983082427*${}^{\wedge}$-32}-\text{4.5948958062512910951997009159524637155472195553235558338875204958$\grave{ }$29.944488042052914*${}^{\wedge}$-32} i$$ and for the right hand side:
$$-\text{8.1747358863640979486289267057810389747002120618563412538892819063$\grave{ }$30.082847849900105*${}^{\wedge}$-32}-\text{4.9430723869496501619456415921898782802016090061312604239413643251$\grave{ }$29.864371090122273*${}^{\wedge}$-32} i$$ which are different.

Mathematica:

Clear[f, A, B, n, s, a, b, x, m];
f[x_] := Zeta[x];
A[n_, s_] := 
 Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/f[s + k/n - 1/n], {k, 1, n}]
B[n_, s_] := 
 Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/f[s + k/n], {k, 1, n}]
X[n_, s_] := 
 Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/f[-s + k/n - 1/n], {k, 1, n}]
Y[n_, s_] := 
 Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/f[-s + k/n], {k, 1, n}]

n = 30;
s = 1/3 + 14*I;
N[A[n, s]*X[n, s], 30]
N[B[n, s]*Y[n, s], 30]
Mats Granvik
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