This question is in reference to an answer I posted here yesterday.
In it I derived the partition function for a harmonic oscillator as follows
$$q = \sum_{j}e^{-\frac{\epsilon_j}{kT}}$$
For the harmonic, oscillator $\epsilon_j = (\frac{1}{2}+j)\hbar \omega$ for $j \in \{ 0,1,2.. \}$
Note that $\epsilon_0 \neq 0$ there exists a zero point energy.
Let's write out a few terms $$q = e^{-\frac{\hbar \omega/2}{kT}} + e^{-\frac{\hbar \omega3/2}{kT}} + e^{-\frac{\hbar \omega5/2}{kT}} +..... $$
factoring out $e^{-\frac{\hbar \omega/2}{kT}}$
$$q = e^{-\frac{\hbar \omega/2}{kT}} \left( 1+ e^{-\frac{\hbar \omega}{kT}} + e^{-\frac{2\hbar \omega}{kT}} +.....\right) $$
The sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \frac{1}{(1-x)} $$ herein, $ x \equiv e^{-\frac{\hbar\omega}{kT}} $
Putting all of this together
$$q = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}$$
I happened to check Atkins Physical chemistry (10th edition) for the same derivation earlier today.
On page 620, the vibrational partition function using the harmonic oscillator approximation is given as $$q = \frac{1}{1-e^{-\beta h c \nu'}}$$, $\beta$ is $\frac{1}{kT}$ and $\nu'$ is wave number
This result was derived in brief illustration 15B.1 on page 613 using a uniform ladder. However, in that illustration the uniform ladder starts at 0, but the harmonic oscillator has a zero point energy (which I accounted for in my derivation).
I discussed this with my instructor and he pointed out that as $T \rightarrow 0$ $q \rightarrow 1 $ (since only one state is thermally accessible.
The result derived in Atkins' does that indeed, but mine goes $q \rightarrow 0$.
Now, in the formalism developed in Atkins the do set the ground state energies to zero (on page 605), and basically add non-zero ground state energies to a calculated $\langle \epsilon \rangle$, which is fine.
Anyway, I would love it if someone could weigh in on this and help me resolve this conceptual mess in my head.