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The trace can be calculated in any basis. Therefore, we will calculate it in the oscillator eigenbasis $\{|n\rangle\}$ as follows. First, $$\text{Tr}\left( e^{-H/ k_B T} \right) = \sum_{n=0}^{\infty}\langle n|e^{-H/ k_B T}|n\rangle =\sum_{n=0}^{\infty}\langle n|e^{-\hbar\omega(n+1/2)/ k_B T}|n\rangle =\sum_{n=0}^{\infty}e^{-\hbar\omega(n+1/2)/ k_B T}. $$ This last sum can be evaluated because it can be written as a geometric series: $$\text{Tr}\left( e^{-H/ k_B T} \right) =e^{-\hbar\omega/2k_B T}\sum_{n=0}^{\infty}\left(e^{-\hbar\omega / k_B T}\right)^n =e^{-\hbar\omega/2k_B T}\frac{1}{1-e^{-\hbar\omega / k_B T}} =\frac{e^{\hbar\omega / 2k_B T}}{e^{\hbar\omega / k_B T}-1} $$ Then, note that $$\text{Tr}(\rho H) = \text{Tr}\left(\frac{e^{-H / k_B T}}{\text{Tr}\left( e^{-H/ k_B T} \right) } H\right) =\frac{\text{Tr}\left(e^{-H / k_B T}H\right)}{\text{Tr}\left( e^{-H/ k_B T} \right) }. $$ So, we have to do the top: \begin{align*} \text{Tr}\left(e^{-H / k_B T}H\right) &=\sum_{n=0}^{\infty}\langle n|e^{-H / k_B T}H|n\rangle =\sum_{n=0}^{\infty}e^{-\hbar\omega(n+1/2)/ k_B T}\hbar\omega\left(n+\frac{1}{2}\right)\\ &=\frac{\hbar\omega}{2}\sum_{n=0}^{\infty}e^{-\hbar\omega(n+1/2)/ k_B T} + \hbar\omega\sum_{n=0}^{\infty}e^{-\hbar\omega(n+1/2)/ k_B T}n\\ &=\frac{\hbar\omega}{2}\frac{e^{\hbar\omega / 2k_B T}}{e^{\hbar\omega / k_B T}-1} + \hbar\omega e^{-\hbar\omega / 2k_B T} \frac{e^{\hbar\omega / 2k_B T}}{e^{\hbar\omega / k_B T}-1} \frac{1}{e^{\hbar\omega / k_B T}-1} \end{align*} Putting these together yields $$\text{Tr} (\rho H) = \frac{\hbar \omega}{2} + \frac{\hbar \omega}{e^{\hbar \omega / k_B T} - 1} \, .$$