I'm having trouble with this question.
$$E=\sum_0^\infty \frac{nhve^{-nx}}{\sum_0^\infty e^{-nx}}$$
where x=(hv)/(kT)
Evaluate both sums and show that E=hv(e^x -1)^-1
I've tried comparing \sum_0^/infty${e^(-nx)}$ to geometric series and expanding the sums but I am unsure how to continue. Any hints would be appreciated.
Outline: Let $f(t)=\sum_0^\infty e^{-nt}$. Note this is a geometric series with first term $1$ and common ratio $e^{-t}$. If $t$ is positive it has sum $\frac{1}{1-e^{-t}}$.
Differentiate the series for $f(t)$ term by term. We get $\sum_0^\infty -ne^{-nt}$. So by differentiating $\frac{1}{1-e^{-t}}$ we can get an expression for the sum on top.
Finally, substitute the known value of $x$ for $t$.