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

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


which returns "True" for each zeta zero I have tried.

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.