= Solution
Write $K=\beta J$, where $\beta=1/(k_BT)$ is the <inverse temperature>, $T$ is <temperature> and $k_B$ is the <Boltzmann constant>. In the <canonical ensemble>, each <spin configuration> has <probability> $e^{-\beta H}/Z_N$. The nearest-neighbor <Boltzmann factors> multiply, so the <spin-chain transfer matrix>, in the order $+1,-1$, is
$$
\mathsf T_{ss'}=e^{-Kss'},\qquad
\mathsf T=\begin{pmatrix}e^{-K}&e^K\\e^K&e^{-K}\end{pmatrix}.
$$
Multiplying its entries around the ring and summing over every <Ising spin> yields the <partition function>:
$$
Z_N=\sum_{S_1,\ldots,S_N}\mathsf T_{S_1S_2}\cdots\mathsf T_{S_NS_1}
=\operatorname{tr}\mathsf T^N.
$$
The <trace> implements the <periodic boundary conditions>; replacing it by an endpoint sum would describe a different <partition function>. The normalized <eigenvectors> and corresponding <eigenvalues> are
$$
|+\rangle=\frac1{\sqrt2}\binom11,\quad
|-\rangle=\frac1{\sqrt2}\binom1{-1},\qquad
\lambda_+=2\cosh K,\quad \lambda_-=-2\sinh K.
$$
The negative <eigenvalue> is essential for the <antiferromagnetic Ising model>. Introduce their ratio
$$
t=\frac{\lambda_-}{\lambda_+}=-\tanh K.
$$
The exact finite-$N$ <partition function> and <Helmholtz free energy> are
$$
\boxed{Z_N=(2\cosh K)^N+( -2\sinh K)^N
=(2\cosh K)^N(1+t^N),}
$$
$$
\boxed{F_N=-\frac1\beta\log Z_N
=-\frac N\beta\log(2\cosh K)-\frac1\beta\log(1+t^N).}
$$
For finite $K>0$, $|t|<1$, so $Z_N>0$ for both even and odd $N$. The last term retains the finite-ring effect of <geometric frustration>.
To insert an <Ising spin> into this <trace>, use the <diagonal matrix> $\mathsf S=\operatorname{diag}(1,-1)$. In the <eigenbasis> of $\mathsf T$, it exchanges $|+\rangle$ and $|-\rangle$, and hence has vanishing diagonal entries. The <one-point correlation function> is therefore
$$
\boxed{\langle S_1\rangle=
\frac{\operatorname{tr}(\mathsf S\mathsf T^N)}{Z_N}=0.}
$$
The same conclusion follows directly from <spin inversion symmetry>: reversing every <Ising spin> preserves the <energy> and the <Boltzmann factor> while reversing $S_1$.
For the <two-point correlation function>, the two insertions divide the ring into paths of $n$ and $N-n$ bonds. Thus
$$
C_n\equiv\langle S_1S_{n+1}\rangle
=\frac{\operatorname{tr}(\mathsf S\mathsf T^n\mathsf S\mathsf T^{N-n})}{Z_N}
=\frac{\lambda_-^n\lambda_+^{N-n}+\lambda_+^n\lambda_-^{N-n}}
{\lambda_+^N+\lambda_-^N}.
$$
In particular, the exact <finite-ring antiferromagnetic Ising correlation> is
$$
\boxed{C_n=\frac{t^n+t^{N-n}}{1+t^N},\qquad t=-\tanh(\beta J),\quad0\le n\le N-1.}
$$
Because the <one-point correlation functions> vanish, this is also the <connected correlation function>. It gives $C_0=1$ and $C_n=C_{N-n}$, as required by $S_1^2=1$ and exchanging the two paths around the ring. At $K=0$, distinct <Ising spins> are independent and $C_n=0$ for $1\le n\le N-1$; $C_0=1$ is understood separately if a power $0^0$ occurs in the displayed expression.
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