= Solution
Share each on-site field energy equally between its two neighboring bonds. In the ordered basis $\sigma=+1,-1$, the symmetric <transfer matrix for a classical spin chain> is
$$
\boxed{T_{ss'}=e^{-\beta Jss'+\frac12\beta B(s+s')},\qquad
T=\begin{pmatrix}e^{-K+h}&e^K\\e^K&e^{-K-h}\end{pmatrix},\quad K=\beta J,\ h=\beta B.}
$$
The product $\prod_jT_{\sigma_j\sigma_{j+1}}$ recovers each bond and each field term once. Summing the periodic indices closes a <trace>, giving $Z_N=\operatorname{Tr}T^N$. The two <eigenvalues> are
$$
\boxed{\lambda_\pm=e^{-K}\cosh h
\pm\sqrt{e^{-2K}\sinh^2h+e^{2K}},\qquad Z_N=\lambda_+^N+\lambda_-^N.}
$$
The square-root argument is positive. Moreover $\det T=e^{-2K}-e^{2K}<0$, so $\lambda_-<0$ even at nonzero field. Its sign must not be discarded merely because all matrix entries are positive. The <Perron–Frobenius theorem> ensures that the positive <eigenvalue> $\lambda_+$ has the larger modulus, and the infinite-chain <free energy> is $f_\infty=-\beta^{-1}\log\lambda_+$. For even $N$, both terms in the finite-size <partition function> are positive.
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