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I'm asking my question here and not in physics SE for the fact that, even though the result comes from physics, the problem I'm having is more mathematical.

Calculating the heat capacity for roto-vibrational state of a molecule, I arrived at this series for the partition function of the rotational states

$$ Z(\beta) = \sum_k (2k+1)e^{-\beta Bk(k+1)}$$

from the top of my head I don't think that there's a closed form for this infinite series (if there's one I would gladly accept it). So I wanted to know if there's something I can say about it. Physically speaking the rotational energy $Bk(k+1)$ is small wrt to $\beta = 1/k_b T$ for temperatures over room temperatures (approx $300$ K).

How can I approximate that sum so to arrive at some closed form for the heat capacity?

$$C_v = -\beta^2\frac{\partial^2}{\partial\beta^2}\log Z(\beta)$$

I don't even think that the series of the derivatives uniformly converges so to put the derivative in the series.

So in general my question is

What useful things can I say about this series so to gain some insights on the heat capacity?

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Unfortunately, rather than a closed form you'll have to settle for a famously physicist-approved approximation of this series as an integral. With $f(k):=k^2+k$,$$Z=\sum_{k\ge 0}f^\prime(k)\exp(-\beta Bf(k))\approx\int_0^\infty f^\prime(k)\exp(-\beta Bf(k))dk=-\frac{1}{\beta B}[\exp(-\beta Bf(k))]_0^\infty=\frac{1}{\beta B}$$(provided your constants are positive). If I may deviate slightly from your definition of $C_v$, the mean energy is $-\partial_\beta\ln Z\approx\frac{1}{\beta}=k_BT$, while $C_v$ is the $T$-derivative of that, i.e. approximately $k_B$.

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  • $\begingroup$ That's how I was thinking to do it, clearly the result is somewhat coherent with the theory, so I'll gladly accept it. Sadly, my quest was on how to, if it was possible, get a more exact answer rather then the integral approximation! But I guess that's the best we can do, thank you! $\endgroup$
    – Quiver
    May 19, 2019 at 9:20

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