The limit on the ratio of the Dirichlet eta functions If we accept--  $s_o$ --  as one of the non-trivial  zeros of the Riemann zeta function  by $0 <Re(s_o)<1$   and     $Re(s_o)$ is the real part of a complex variable, we know: 
$$\eta(s_o ) = \sum_{1}^\infty \frac {(-1)^{n-1}} {n^{s_o}}=0 .... and.... \eta(1-s_o ) = \sum_{1}^\infty \frac {(-1)^{n-1}} {n^{1-s_o}}=0  \tag{1}$$ 
Now, let us accept here that--  $s_x$--  refers the values which are different than the non-trivial zeros of the Riemann zeta function. We mean  $s_x≠s_{∘}$, thus we see   $\eta(s_{x})≠0$  and  $\eta(1-s_{x})≠0$.
Now, let us define $F(s)$ function as below: 
$$F(s_x)=\frac {\sum_{1}^{\infty}\frac {(-1)^{n-1}} {n^{s_x}}} {\sum_{1}^\infty \frac {(-1)^{n-1}} {n^{1-s_x}}} \tag{2}$$ 
QUESTION: Can we reach the last equation (5) in order? I mean do I make any mistake to obtain it here?  
$$F(s_x)=\frac {\lim_{m\rightarrow\infty}\sum_{1}^{m}\frac {(-1)^{n-1}} {n^{s_x}}} {\lim_{m\rightarrow\infty}\sum_{1}^{m} \frac {(-1)^{n-1}} {n^{1-s_x}}} \tag{3}$$ 
$$F(s_x)=\lim_{m\rightarrow\infty}\frac {\sum_{1}^{m}\frac {(-1)^{n-1}} {n^{s_x}}} {\sum_{1}^{m} \frac {(-1)^{n-1}} {n^{1-s_x}}} \tag{4}$$
Then, we can write it with the non-trivial  zeros under the following limit--(${m\rightarrow\infty}$)
$$F(s_o)=\lim_{m\rightarrow\infty}\frac {\sum_{1}^{m}\frac {(-1)^{n-1}} {n^{s_o}}} {\sum_{1}^{m} \frac {(-1)^{n-1}} {n^{1-s_o}}} \tag{5}$$
 A: *

*See my answers to your previous questions for $\Re(s) > 0$
$$\sum_{n=1}^N (-1)^{n+1} n^{-s} = \eta(s) + \frac{(-1)^N}{2} (2N)^{-s}+O(N^{-s-1} \frac{s(s+1)}{\Re(s)})$$
Do you see how it answers thisone too. 

*The functional equation is the equality of meromorphic functions which is valid pointwise in the critical strip
$$\eta(s) = \phi(s) \eta(1-s), \qquad \phi(s) =\frac{1-2^{1-s}}{1-2^{s} } \chi(s), \chi(s) = 2^{s} \pi^{s-1} \Gamma(1-s) \sin(\pi s/2) $$ 


*

*Thus for $\Re(s) \in (0,1)$ $$\lim_{N \to \infty} \frac{\sum_{n=1}^N (-1)^{n+1} n^{-s}}{\sum_{n=1}^N (-1)^{n+1} n^{s-1}}$$ converges to $\phi(s)$ iff $\eta(s) \ne 0$. 

*If $\eta(s) = 0$ : or $\Re(s) \le 1/2$ and it diverges, or $\Re(s) > 1/2$ and it converges to $0$.
Thus the RH is true iff for every $\Re(s) \in (0,1)$ $$\lim_{N \to \infty} \frac{1}{\log N}\frac{\sum_{n=1}^N (-1)^{n+1} n^{-s}}{\sum_{n=1}^N (-1)^{n+1} n^{s-1}}= 0$$
Of course there is no elementary way to show it, because an elementary argument would also apply to the Dirichlet series with functional equation but no Euler product, for which the RH fails.

*If $\eta(s) = 0$ and $s_N \to s$ you can also find if $\lim_{N \to \infty} \frac{\sum_{n=1}^N (-1)^{n+1} n^{-s_N}}{\sum_{n=1}^N (-1)^{n+1} n^{s_N-1}}$ converges or not and to what value, depending on $\Re(s_N)$ and the rate of convergence of $s_N$
