I don't understand something about the residues of $\zeta'(s)/\zeta(s)$.
I'm studying complex analysis using Stein's Complex Analysis. The zeros of the Riemann zeta function are $-2n$ ($n$ is natural) and the non-trivial zeros.
Zeros of the Riemann zeta function aren't compose a limit point, the domain is a connected set.
Therefore, is residue of $\zeta'(s)/\zeta(s)$ order of zero of $\zeta(s)$?
Residues of $\frac{x^s}{s}\frac{\zeta'(s)}{\zeta\phantom{'}(s)}$
This article answer is $$\lim_{s\to\rho} (s-\rho)\zeta'(s)/\zeta(s) = 1$$
if that answer is true, the order of non-trivial zero is $1$.
But the simple zero hypothesis ("all zero of Riemann zeta function are simple") is unknown.
I dont understand this.
This has nothing specific to do with the zeta-function. When $f(s)$ is an arbitrary meromorphic function besides $0$, the logarithmic derivative $f'(s)/f(s)$ at any point $a$ in its domain has a pole of order at most $1$ with ${\rm Res}_{s=a}(f'/f) = {\rm ord}_{s=a}(f)$: the residue of $f'/f$ at a point is the order of vanishing of $f$ that point (positive, negative, or zero).
This comes from looking at the Laurent expansion of $f$ at $a$ and seeing what that tells you about the leading term of $f'/f$ at $a$: if $f(s) = c(s-a)^m + \cdots$ near $a$, where $c \not= 0$, then $f'(s) = mc(s-a)^{m-1} + \cdots$ near $a$. Show $(f'/f)(s) = m/(s-a) + \cdots$, where $\cdots$ in all cases means higher-order terms. The residue of $f'/f$ at $a$ is $m$, which is the order of vanishing of $f$ at $a$.