Pole of Riemann zeta and Riemann zeta zeros, prove this relation. Prove this relation:
$$\displaystyle \lim_{s\to 1} \, \left(\zeta (s)-\frac{\zeta '(s-1+\rho _n)}{\zeta \left(s-1+\rho _n\right)}\right)=\gamma -\frac{\zeta ''(\rho _n)}{2 \zeta '(\rho _n)}\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
Where $\rho _n$ is a Riemann zeta zero - either non-trivial or trivial, $\zeta(s)$ is the Riemann zeta function, $\gamma = 0.57721566490...$ is the Euler Mascheroni constant and $\zeta'(s)$ is a derivative.
As often with my questions they might be trivial to answer, but I really don't know where to start. I found this by changing equation $(4)$ in this question:
$$-\frac{\zeta'(s)}{\zeta(s)}=\sum\limits_{k=1}^{\infty} \frac{\Lambda(k)}{k^s} = \sum\limits_{n=1}^{\infty} \left(\frac{1}{n}\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}-\frac{1}{n}\right)\;\;\;\;\;\;\;\;\;\;\;\;(4)$$
where $\Lambda(k)$ is the von Mangoldt function, $\mu(d)$ is the Möbius function and $d$ is a divisor.
Some Mathematica code to demonstrate equation $(1)$ above:
Table[N[Limit[
    Zeta[s] - 
     Derivative[1][Zeta][s - 1 + ZetaZero[n]]/
      Zeta[s - 1 + ZetaZero[n]], s -> 1] == 
       EulerGamma - 
    Derivative[2][Zeta][
      ZetaZero[n]]/(2*Derivative[1][Zeta][ZetaZero[n]])], {n, 1, 12}]

Table[N[Limit[
    Zeta[s] - Derivative[1][Zeta][s - 1 - 2*n]/Zeta[s - 1 - 2*n], 
    s -> 1] == 
       EulerGamma - 
    Derivative[2][Zeta][-2*n]/(2*Derivative[1][Zeta][-2*n])], {n, 1, 
  12}]

which returns "True" for each zeta zero I have tried.
 A: Let me write $z = s-1$ for a little more convenient notation in the expansions.
From the pole of $\zeta$ we have the Laurent expansion
$$\zeta(1+z) = \frac{1}{z} + \gamma + O(z).$$
Taylor expansion of $\zeta$ and $\zeta'$ around $\rho_n$ yields
$$\begin{align}
\frac{\zeta'(z+\rho_n)}{\zeta(z+\rho_n)}
&= \frac{\zeta'(\rho_n) + \zeta''(\rho_n)z + O(z^2)}{\zeta'(\rho_n)z + \frac{1}{2}\zeta''(\rho_n)z^2 + O(z^3)}\\
&= \frac{1}{z}\cdot \frac{1 + \frac{\zeta''(\rho_n)}{\zeta'(\rho_n)}z + O(z^2)}{1 + \frac{\zeta''(\rho_n)}{2\zeta'(\rho_n)}z + O(z^2)}\\
&= \frac{1}{z}\left(1 + \frac{\zeta''(\rho_n)}{\zeta'(\rho_n)}z + O(z^2)\right)\left(1 - \frac{\zeta''(\rho_n)}{2\zeta'(\rho_n)}z + O(z^2)\right)\\
&= \frac{1}{z}\left(1 + \frac{\zeta''(\rho_n)}{2\zeta'(\rho_n)}z + O(z^2)\right),
\end{align}$$
so
$$\zeta(1+z) - \frac{\zeta'(z+\rho_n)}{\zeta(z+\rho_n)} = \gamma + \frac{1}{z} - \frac{1}{z}\left(1 + \frac{\zeta''(\rho_n)}{2\zeta'(\rho_n)}z\right) + O(z) = \gamma - \frac{\zeta''(\rho_n)}{2\zeta'(\rho_n)} + O(z)$$
and the limit follows.
