= Solution
Take the <thermodynamic limit> $N\to\infty$ at fixed separation $n$ and fixed finite $K=\beta J>0$. Since $|t|=\tanh K<1$, the terms $t^{N-n}$ and $t^N$ in the <finite-ring antiferromagnetic Ising correlation> vanish, through either even or odd $N$. Therefore
$$
C_n\longrightarrow t^n=(-1)^n(\tanh K)^n
=(-1)^n\exp\bigl[n\log(\tanh K)\bigr].
$$
Matching the decaying <exponential function> gives the <correlation length>, in units of the site spacing:
$$
\boxed{\langle S_1S_{n+1}\rangle=(-1)^ne^{-n/\xi},\qquad
\xi=-\frac1{\log\tanh(\beta J)}
=\frac1{\log\coth(\beta J)}.}
$$
If the physical spacing is $a$, the physical <correlation length> is $a\xi$. The alternating sign records the preferred opposite orientation of neighboring <Ising spins>; the finite <correlation length> shows that this preference does not maintain long-range <Néel order> at positive <temperature>.
For large $K$, $\tanh K=1-2e^{-2K}+O(e^{-4K})$, so $\xi\sim e^{2K}/2$: the <correlation length> becomes large on cooling, but diverges only at zero <temperature>. At infinite <temperature>, $K\to0^+$ and $\xi\to0$; the separate equal-site <two-point correlation function> remains $C_0=1$.
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