I therefore believe that the Riemann hypothesis is equivalent to showing:
$$\lim_{n\to \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-\frac{1}{n}\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s\right)}}}+s\right)=\lim_{n\to \infty } \, \left(\frac{1}{1-\frac{\sum _{k=1}^n \frac{(-1)^{k-1} \left( \begin{array}{c} n-1 \\ k-1 \end{array} \right)}{\zeta \left(\frac{k}{n}+s\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \left( \begin{array}{c} n-1 \\ k-1 \end{array} \right)}{\zeta \left(\frac{k}{n}+s-…