= Solution
Set $a=2mL^2/(\hbar^2\beta)$ and $C=\sqrt{m/(2\pi\hbar^2\beta)}$. For the reflected images, the intervals $rL-q$, $0\le q\le L$, tile the real line once. A <Gaussian integral> gives
$$
C\sum_r\int_0^Le^{-2m(rL-q)^2/(\hbar^2\beta)}dq=C\int_{-\infty}^\infty e^{-2mx^2/(\hbar^2\beta)}dx=\frac12.
$$
Consequently $Z=CL\sum_re^{-ar^2}-1/2$. The <Poisson summation formula> applied to the Gaussian yields
$$
\sum_{r\in\mathbb Z}e^{-ar^2}=\sqrt{\frac\pi a}\sum_{n\in\mathbb Z}e^{-\pi^2n^2/a},\qquad CL\sqrt{\frac\pi a}=\frac12.
$$
The zero dual mode cancels the reflected contribution, while the positive and negative modes pair:
$$
\boxed{Z=\frac12\sum_{n\in\mathbb Z}e^{-\beta\hbar^2\pi^2n^2/(2mL^2)}-\frac12=\sum_{n=1}^\infty e^{-\beta\hbar^2\pi^2n^2/(2mL^2)}}.
$$
This proves equality with the <energy eigenstate> calculation. The image expansion converges rapidly at small $\beta$; the spectral expansion converges rapidly at large $\beta$.
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