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Given the logarithmic derivative of the zeta function $\dfrac{\zeta^\prime (s)}{\zeta(s)}$ how does it behave near $s=1$?

I mean if for $s=1$ the Laurent series for the logarithmic derivative becomes

$$\frac{\zeta^\prime (s)}{\zeta(s)}= A+ (s-1)^{-1}$$

where $A$ is a real number constant.

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    The $A$ you have there is the Euler-Mascheroni constant, $\gamma$. Actually, your series goes like $\frac{\zeta^\prime (s)}{\zeta(s)}=-\frac1{s-1}+\gamma-(\gamma^2+2\gamma_1)(s-1)+\cdots$ where $\gamma_1$ is the first Stieltjes constant.2011-10-09
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    Not directly related but the following is associated with a symmetric matrix: $$-\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}-\zeta (c)\right)$$ https://oeis.org/A1918982016-07-17
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    @Mats where can I find a proof of that statement?2017-07-29
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    @kaffeeauf Starting from the mentioned matrix to arrive at this generating function I don't know how to prove the generating function for it. However the truth of the statement: $$-\frac{\zeta '(s)}{\zeta (s)}=\lim_{c\to 1} \, \left(\frac{\zeta (c) \zeta (s)}{\zeta (c+s-1)}-\zeta (c)\right)$$ I have been told by I think it was user Lucia at mathoverflow, that this is standard definition of the derivative somehow. Also, Mathematica knows that this statement is true: Limit[Zeta[s]*Zeta[c]/Zeta[s + c - 1] - Zeta[c], c -> 1] The answer at mathoverflow got down voted so I deleted it.2017-07-29

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