= Solution
The frequency-independent measure factor cancels in the ratio:
$$
R_N=\frac{\omega_2}{\omega_1}\prod_{r=1}^N\frac{1+(\beta\omega_2/(2\pi r))^2}{1+(\beta\omega_1/(2\pi r))^2}.
$$
Upon <analytic continuation> in frequency, this has <poles> at $\omega_1=0,\ \pm2\pi i r/\beta$ and zeros at $\omega_2=0,\ \pm2\pi i r/\beta$, for $1\leq r\leq N$, apart from cancellations when numerator and denominator vanish together. They match those of $\sinh(\beta\omega_2/2)/\sinh(\beta\omega_1/2)$. To justify equality, rather than merely matching this divisor, use the <hyperbolic-sine infinite product>
$$
\frac{\sinh(\beta\omega/2)}{\beta\omega/2}=\prod_{r=1}^\infty\left[1+\left(\frac{\beta\omega}{2\pi r}\right)^2\right].
$$
It follows from pairing the <Weierstrass product for the reciprocal gamma function> at opposite imaginary arguments and using the <Gamma reflection formula>. Normal convergence on compact sets follows from $\sum r^{-2}<\infty$. Thus, for positive real frequencies,
$$
\boxed{\lim_{N\to\infty}R_N=\frac{\sinh(\beta\omega_2/2)}{\sinh(\beta\omega_1/2)}=\frac{\mathcal Z(\omega_1,\beta)}{\mathcal Z(\omega_2,\beta)}.}
$$
The cutoff ratio determines the frequency dependence, agreeing with the <thermal partition function of a quantum harmonic oscillator>.
The printed assertion about an absolute $\lim_N\mathcal Z_N$ needs a normalization qualification: independence of $A_N$ from $\omega$ alone does not ensure existence of a nonzero limit. Set $B_N=A_N\prod_{r=1}^N\nu_r^{-2}$. Then $\mathcal Z_N=B_N\omega^{-1}\prod_r(1+\omega^2/\nu_r^2)^{-1}$, so if $B_N\to B\ne0$, its limit is $B\beta/[2\sinh(\beta\omega/2)]$. The choice $B_N=1/\beta$ gives the required limit. Conversely, $B_N=(1+(-1)^N/2)/\beta$ is a positive frequency-independent normalization with no absolute limit, though every ratio above still converges. Matching zeros and poles alone also permits nonvanishing <entire functions>; the convergent normalized product is what rules them out here.
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