= Solution
Put $x=e^{-\beta J}$. Factoring out the <ground state> <Boltzmann factor> from the exact <canonical partition function> gives
$$
Z=e^{\beta JL}\left\{[1+(Q-1)x]^L+(Q-1)(1-x)^L\right\}.
$$
At fixed finite $L$, the <binomial theorem> gives
$$
\begin{aligned}
[1+(Q-1)x]^L+(Q-1)(1-x)^L
&=1+(Q-1)\\
&\quad+[L(Q-1)-(Q-1)L]x\\
&\quad+\binom L2[(Q-1)^2+Q-1]x^2+O_{L,Q}(x^3)\\
&=Q\left[1+\binom L2(Q-1)x^2+O_{L,Q}(x^3)\right].
\end{aligned}
$$
Taking the <logarithm> yields the <finite-ring low-temperature expansion> of the <Helmholtz free energy>,
$$
\boxed{F=-JL-\frac{\log Q}{\beta}
-\frac{L(L-1)(Q-1)}{2\beta}e^{-2\beta J}
+O_{L,Q}\!\left(\frac{e^{-3\beta J}}{\beta}\right).}
$$
The constant term is the <ground state> <energy>. The term $-\beta^{-1}\log Q$ is the contribution of the <ground state> <degeneracy factor>. The first thermal correction follows directly by retaining the <first excited state> manifold:
$$
Z=e^{-\beta E_0}\left[d_0+d_1e^{-2\beta J}+\cdots\right],
\qquad \frac{d_1}{d_0}=\binom L2(Q-1).
$$
Each pair of <domain walls> costs $2J$, explaining the absence of an $e^{-\beta J}$ term. For $Q=2$, every configuration has an even number of <domain walls>, so the displayed remainder improves to $O_L(e^{-4\beta J}/\beta)$.
This is an expansion for fixed $L$, not a uniform <thermodynamic limit>. In particular, the small correction requires $\binom L2(Q-1)x^2\ll1$, with higher terms also controlled as $x\to0$ at fixed $L,Q$. Taking $L\to\infty$ first instead selects $\lambda_+$ and gives
$$
\lim_{L\to\infty}\frac FL=-J-\frac1\beta\log[1+(Q-1)x].
$$
Its leading correction is of order $x$. The finite-ring pair constraint and the extensive population of <domain walls> in an arbitrarily long chain explain the different ordering of these two limits.
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