How does the average energy from equation (41.15) of Feynman's lectures approach $kT$ as omega goes to zero? Feynman said the following equation should approach $kT$ as $\omega$ goes to zero, or $T$ goes to infinity.
$$\langle E \rangle = \frac{\hbar \omega}{e^{\hbar \omega/kT}-1}$$
Does anybody know how to prove this?
Get use of L'Hopital's rule where ' stands for the derivative
$$\lim_{\omega \to 0} \langle E\rangle= \lim_{\omega \to 0}\frac{(\hbar \omega)'}{(e^{\frac{\hbar \omega}{kT}}-1)'}  = \lim_{\omega \to 0} kT e^{-{\frac{\hbar \omega}{kT}}}=kT$$
Use the Taylor series for the exponential function,
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots.$$
In addition to giving you the $\omega\to 0$ limit, it will give you the low-frequency corrections.