How were these calculated? $$\langle E\rangle = \frac{\displaystyle\int_0^\infty Ee^{-E/kT}{\rm d}E}{\displaystyle\int_0^{\infty}e^{-E/kT}{\rm d}E}=kT$$ $$ \langle E\rangle = \frac{\displaystyle\sum_{n=0}^{\infty}nh\nu e^{-nh\nu/kT}}{\displaystyle\sum_{n=0}^\infty e^{-nh\nu/kT}}=\frac{\displaystyle h\nu}{\displaystyle e^{h\nu/kT}-1}$$ $$\int_0^\infty \frac{8\pi \nu^2}{c^3}.\frac{\displaystyle h\nu}{\displaystyle e^{h\nu/kT}-1}{\rm d}\nu=\frac 4c\sigma T^4$$ $$\sigma:=\frac{2\pi^5k^4}{15h^3c^2}$$
I did the first one using integration by parts, numerator is $(kT)^2$ and denominator $(kT)$ using $\int u{\rm d}v=uv-\int v{\rm d}u$. Also I know $\displaystyle\sum_{n=0}^{\infty}x=\frac1{1-x},|x|<1$