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]

Mathematica again:

    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 = 120;
    s = N[1 + 2*I , 200];
    Block[{$MaxExtraPrecision = 600}, N[A[n, s]*X[n, s]/(B[n, s]*Y[n, s]), 20]]
    N[1 + 5/((s/I)^2 + 4), 20]

which numerically suggests that:

$$\lim_{n\to \infty } \, \left(\frac{A(n,s) X(n,s)}{B(n,s) Y(n,s)} \right) =1+\frac{5}{4+\left(\frac{s}{i}\right)^2} $$
for those $s$ closest to the first trivial zero.