# Evaluation of Planck's sum

To evaluate Planck's sum
$$\langle E \rangle = {\sum_{n = 0}^\infty n h \nu \exp\bigl({ - n h \nu \over k T}\bigr) \over \sum_{n = 0}^\infty \exp\bigl({ - n h \nu \over k T}\bigr)}~,$$
for the average energy per mode$\langle E \rangle$, it is convenient to introduce the variable
$$\alpha \equiv {1 \over kT}$$
so $$\langle E \rangle = {\sum_{n = 0}^\infty n h \nu \exp(- \alpha n h \nu) \over \sum_{n = 0}^\infty \exp(- \alpha n h \nu)}~.$$

Next consider the quantity
$$- {d \over d\alpha}\biggl[\ln \sum_{n = 0}^\infty \exp(- \alpha n h \nu)\biggr]$$
Using the chain rule to take the derivative yields
$$- {d \over d\alpha}\biggl[\ln \sum_{n = 0}^\infty \exp(- \alpha n h \nu)\biggr] = {1 \over \sum_{n = 0}^\infty \exp(- \alpha n h \nu)} {d \over d \alpha} \biggl[\sum_{n = 0}^\infty \exp(-\alpha n h \nu)\biggr]$$
$$- {d \over d\alpha}\biggl[\ln \sum_{n = 0}^\infty \exp(- \alpha n h \nu)\biggr] = {\sum_{n = 0}^\infty n h \nu \exp (- \alpha n h \nu) \over \sum_{n = 0}^\infty \exp(- \alpha n h \nu)}$$
Thus
$$\langle E \rangle = - {d \over d\alpha}\biggl[\ln \sum_{n = 0}^\infty \exp(- \alpha n h \nu)\biggr]~.$$
The expanded sum
$$\sum_{n = 0}^\infty \exp(- \alpha n h \nu) = 1 + [\exp( - \alpha h \nu)]^1 + [\exp( - \alpha h \nu)]^2 + \dots ~.$$
has the form
$$1 + x + x^2 + \dots = (1 - x)^{-1}~,$$
so
$$\sum_{n = 0}^\infty \exp(- \alpha n h \nu) = [1 - \exp (- \alpha h \nu)]^{-1}$$
and
$$\langle E \rangle = - {d \ln[1 - \exp ( \alpha h \nu)]^{-1} \over d \alpha}~.$$
$$\langle E \rangle = -[1 - \exp( - \alpha h \nu)] (-1) [1 - \exp( - \alpha h \nu)]^{-2} h \nu\exp (- \alpha h \nu)$$
$$\langle E \rangle = { h \nu \exp( - \alpha h \nu) \over 1 - \exp(- \alpha h \nu)}$$
$$\langle E \rangle = { h \nu \over \exp(\alpha h \nu) - 1}$$
$$\langle E \rangle = {h \nu \over \exp\bigl({h \nu \over k T}\bigr) -1}$$