Prove that this limit is $-2$, which is a trivial zero of the Riemann zeta function.

Question

The numerical evaluation of this formula works in Mathematica.

Let: $$s=1$$ and truncate, at some large number, the limit involving the Harmonic numbers: $$f(x)= \lim_{m \rightarrow \infty}H_{m}^{x}$$

Show that:

$$\lim_{n \rightarrow \infty} \left( \left[ 1- \left( \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f(\tfrac{k}{n}+s)} \Bigg/ \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f(\tfrac{k}{n}+s+\tfrac{1}{n})} \right) \right]^{-1} +\frac1n + s \right) = -2$$

which is the trivial Riemann zeta funtion zero nearest to $s=1$.

If one sets $s=1+14i$, one gets an approximation of the first non-trivial Riemann zeta zero.

Mathematica:

(*start*)
(*Mathematica 8.0.1*)
n = 40;(*set n=60 for more digits*)
s = 1;
f[x_] := HarmonicNumber[10^10000, x];
s + 1/n + 
  1/(1 - Sum[(-1)^(k - 1)*Binomial[n - 1, k - 1]/f[s + k/n], {k, 1, 
        n}]/Sum[(-1)^(k - 1)*
        Binomial[n - 1, k - 1]/f[s + k/n + 1/n], {k, 1, n}]);
N[%, n]
(*end*)