 Parity-twisted oscillator thermal trace

For two equal-frequency quantum harmonic oscillators, the ordinary thermal trace is $r/(1-r{)}^2$ and the trace with the parity operator is $r/(1+r{)}^2$, where $r=e^{-\beta \omega }$. Total excitation $N$ has degeneracy $N+1$ and parity $(-1{)}^N$. Therefore the traces expand as ${\sum }_{n\ge 1}(\pm 1{)}^{n-1}n{r}^n$. The alternating weights come from parity insertion in a bosonic state space, not fermionic statistics.