Let:
$s= 1/3 + 14i$
Prove that the real part of this limit converges to $\frac{1}{2}$:

$$\LARGE\Re\left(\lim_{n \rightarrow \infty}\left(\frac{1}{1-\frac{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s+\frac{1}{n}\right)}}}+\frac{1}{n}+s\right)\right)=\frac{1}{2}$$

Mathematica 8.0.1:

n = 100;
s = (1/3 + 14*I);
s + 1/n + 
  1/(1 - Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/Zeta[s + k/n], {k, 1,
         n}]/Sum[(-1)^(k - 1)*
        Binomial[n - 1, k - 1]/Zeta[s + k/n + 1/n], {k, 1, n}]);
N[%, n]

Output:
0.50000000000000000000000000000000000000000000000000000000000000000000
000000000000000000000000000000 +
14.134725141734693790457251983562470270784257115699243175685567460149\ 96342980925676494901039317156101 I

Pari GP:

n = 30
s = (1/2 + 14*I)
s + 1/n + 1/(1 - sum (k = 1, n, (-1)^(k - 1)*binomial(n - 1, k - 1)/zeta (s + k/n))/sum (k = 1, n, (-1)^(k - 1)*binomial(n - 1, k - 1)/zeta (s + k/n + 1/n)))

Copy Paste via mouse button and press Shift Enter in Pari GP to compute.
The precision in Pari GP is not as good as in Mathematica. Therefore $n=30$ instead of $n=100$.