Do plots $(5)$ and $(6)$ go to infinity not at the same rate but at similar rates?

Question

The following has been proven by joriki and GH from MO (see here): assuming that $n>1$, then the von Mangoldt function $$ \Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}} $$ can be computed as the row or column sums of $$ T(n,k) = a(GCD(n,k)). $$ where the Dirichlet inverse of the Euler totient is $$ a(n) = \sum\limits_{d|n} d \cdot \mu(d). $$ The generating function for the matrix $T$ is $$ \sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^s} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{z \rightarrow s} \zeta(z)\sum\limits_{d|n} \frac{\mu(d)}{d^{(z-1)}}}{n^c} = \frac{\zeta(s) \zeta(c)}{\zeta(s + c - 1)}. $$ The generating function for the main diagonal of the matrix $T$ is $$ \sum_{n=1}^{\infty} \frac{T(n,n)}{n^s}=\frac{\zeta (s)}{\zeta (s-1)}. $$ Set $c=s$ and add the generating functions together and take the limit $$ \underset{s\to 1}{\text{lim}}\left(\frac{\zeta (s) \zeta (s)}{\zeta (s+s-1)}+\frac{\zeta (s)}{\zeta (s-1)}\right)=-2 \gamma+2 \log (2 \pi ). $$ But that is not the same as the reciprocal sum over the zeta zeros:
RH Equivalence 5.3, page 6. The Riemann Hypothesis is equivalent to the equality: $$ \sum_{\rho} \frac{1}{|\rho|^2} =\sum_{\rho} \frac{1}{\rho (1{-}\rho)}= 2 + \gamma - \log (4\pi) $$ where the sum is over all complex zeros $\rho = \beta + i\gamma$ of $\zeta(s)$ in the critical strip $0 < \beta < 1$.
However, I notice the following: $$ \underset{s\to 1}{\text{lim}}\left(\frac{\zeta (s) \zeta (s)}{\zeta (s+s-1)}+\frac{\zeta (s)}{\zeta (s-1)}\right)=-2 \gamma+2 \log (2 \pi ) \label{1}\tag{1} $$ $$ \sum_{\rho} \frac{1}{|\rho|^2} =\sum_{\rho} \frac{1}{\rho (1{-}\rho)}= 2 + \gamma - \log (4\pi) \label{2}\tag{2} $$ $$ \sum_{k=1}^{\infty} \frac{1}{-2k(1-(-2k))} = \log (2)-1 \label{3}\tag{3} $$ where \eqref{3} is almost the difference between \eqref{1} and \eqref{2} apart from a sign and a factor of $2$, and the number $1$ which could be interpreted as the pole of zeta.
That is: $$ \underset{\text{factor}}{-2}\left(\underset{\text{non-trivial zeros}}{\sum_{\rho} \frac{1}{\rho (1{-}\rho)}} + \underset{\text{trivial zeros}}{\sum_{k \geq 1} \frac{1}{-2k(1-(-2k))}} - \underset{\text{the pole}}{1}\right)= \underset{s\to 1}{\text{lim}}\left(\underset{\text{the whole matrix}}{\frac{\zeta (s) \zeta (s)}{\zeta (s+s-1)}}+\underset{\text{main diagonal}}{\frac{\zeta (s)}{\zeta (s-1)}}\right) \label{4}\tag{4} $$

Is the relation \eqref{4} just a coincidence or is there something more to it?

Edit 16.09.2024:

The Hadamard product for the Riemann zeta function is:

$$\zeta (s)=\frac{\pi ^{s/2} \prod _{k=1}^{\infty} \left(1-\frac{s}{\rho _k \rho _{-k}}\right)}{2 (s-1) \Gamma \left(\frac{s}{2}+1\right)}$$

Substituting this into:

$$\underset{s\to 1}{\text{lim}}\left(\frac{\zeta (s) \zeta (s)}{\zeta (s+s-1)}+\frac{\zeta (s)}{\zeta (s-1)}\right)$$

and evaluating the limit, we get:

$$-\frac{\left(\prod _{k=1}^n \left(\rho _{-k} \rho _k-1\right)\right) \left(\sum _{k=1}^n \frac{2 \prod _{h=1}^{k-1} \rho _{-h} \rho _h \prod _{k=1}^n \rho _{-k} \rho _k}{\prod _{h=1}^k \rho _{-h} \rho _h}+(\log (4)-4) \prod _{k=1}^n \rho _{-k} \rho _k-\sum _{k=1}^n \frac{2 \prod _{k=1}^n \rho _{-k} \rho _k}{\rho _{-k} \rho _k-1}\right)}{\prod _{k=1}^n \left(\rho _{-k}\right){}^2 \left(\rho _k\right){}^2} \label{5}\tag{5}$$

where $n$ is an integer that goes to infinity. (The latex is too long for the Stack exchange format.)

Then under the Riemann hypothesis that must be equivalent to the formula below:

$$-2 \left(\sum _{k=1}^n \frac{1}{\rho _k \rho _{-k}}+\sum _{k=1}^{\infty } \frac{1}{-2 k (1-(-2 k))}-1\right)\label{6}\tag{6}$$ where $n$ is an integer that goes to infinity.

Mathematica 14 notes:

Clear[nn];
(*nn=79;*) (*nn=Infinity*)
nn = 2;
a = Table[
  limit = -((Product[-1 + ZetaZero[-k]*ZetaZero[k], {k, 1, 
          n}]*((-4 + Log[4])*
           Product[ZetaZero[-k]*ZetaZero[k], {k, 1, n}] + 
               
          Sum[(2*Product[ZetaZero[-k]*ZetaZero[k], {k, 1, n}]*
              Product[ZetaZero[-h]*ZetaZero[h], {h, 1, -1 + k}])/
            Product[ZetaZero[-h]*ZetaZero[h], {h, 1, k}], {k, 1, n}] - 
               
          Sum[(2*Product[ZetaZero[-k]*ZetaZero[k], {k, 1, n}])/(-1 + 
              ZetaZero[-k]*ZetaZero[k]), {k, 1, n}]))/
      Product[ZetaZero[-k]^2*ZetaZero[k]^2, {k, 1, n}]), {n, 1, nn}]
b = Table[
  sum = -2*(Sum[1/(ZetaZero[k]*ZetaZero[-k]), {k, 1, n}] + 
      Sum[1/(-2*k*(1 - (-2*k))), {k, 1, Infinity}] - 1), {n, 1, nn}]
Chop[N[a]]
Chop[N[b]]
f[s_] := 
 Pi^(s/2)*
  Product[1 - s/(ZetaZero[k]*ZetaZero[-k]), {k, 1, nn}]/(2*(s - 1)*
     Gamma[1 + s/2]); Chop[{N[
   Limit[f[s]*f[s]/f[s + s - 1] + (f[s]/f[s - 1]), s -> 1]], 
  N[-2*(Sum[1/(ZetaZero[k]*ZetaZero[-k]), {k, 1, nn}] + 
      Sum[1/(-2*k*(1 - (-2*k))), {k, 1, Infinity}] - 1)]}]

Edit 14.10.2024:

Plotting as a function of $n$ $\eqref{5}$ in gray and $\eqref{6}$ in blue:

Question: Do $\eqref{5}$ in gray and $\eqref{6}$ in blue go to infinity not at the same rate but at similar rates?

Plot of $\eqref{6}$ divided by $\eqref{5}$ as a function of $n$:

Clear[nn];
(*nn=79;*)(*nn=Infinity*)
nn = 79;
Monitor[a = 
   ParallelTable[
    limit = -((Product[-1 + ZetaZero[-k]*ZetaZero[k], {k, 1, 
            n}]*((-4 + Log[4])*
             Product[ZetaZero[-k]*ZetaZero[k], {k, 1, n}] + 
            Sum[(2*Product[ZetaZero[-k]*ZetaZero[k], {k, 1, n}]*
                Product[ZetaZero[-h]*ZetaZero[h], {h, 1, -1 + k}])/
              Product[ZetaZero[-h]*ZetaZero[h], {h, 1, k}], {k, 1, 
              n}] - Sum[(2*
                Product[ZetaZero[-k]*ZetaZero[k], {k, 1, n}])/(-1 + 
                ZetaZero[-k]*ZetaZero[k]), {k, 1, n}]))/
        Product[ZetaZero[-k]^2*ZetaZero[k]^2, {k, 1, n}]), {n, 1, 
     nn}];, n]
Monitor[b = 
   Table[sum = -2*(Sum[1/(ZetaZero[k]*ZetaZero[-k]), {k, 1, n}] + 
        Sum[1/(-2*k*(1 - (-2*k))), {k, 1, Infinity}] - 1), {n, 1, 
     nn}];, n]
ListPlot[Chop[N[a]], PlotStyle -> Gray]
ListPlot[Chop[N[b]]]
Show[%%, %]
ListPlot[Chop[N[b]]/Chop[N[a]]]