Q:

How was this sum simplified to a ratio of Sinh functions?

I came across this simplification in a paper and I do not understand how it was done.
$$\sum_{m=0}^{K-1} \frac{e^{-mT/Kt}(1-e^{-T/Kt})e^{jnw(m+0.5)T/K}} {(1-e^{-T/t})} = \frac{sinh( \frac{T}{2Kt})}{sinh(\frac{T(1-jnwt)}{2Kt})}$$
where $j$ is the imaginary unit, and $w=2\pi/T$
My attempts so far have been to expand the right hand side and get it to match the left hand side using the identity $$sinh(x) = \frac{1-e^{-2x}}{2e^{-x}}$$
However, I cannot find a way to get rid of the summation over $m$.
Ultimately, the LHS simplified to something of the form:
$$A \sum_{m=0}^{K-1} e^{-mT/Kt}e^{jn(m+0.5)2\pi/K}$$
Is it possible to simplify out the summation?

A:

The sum in your final line is a geometric progression:
$$\sum_{m=0}^{K-1} e^{-mT/Kt}e^{jn(m+0.5)2\pi/K}
  =\sum_{m=0}^{K-1} e^{jn\pi/K}\bigl(e^{-T/Kt+jn2\pi/K}\bigr)^m$$
and the sum is
$$e^{jn\pi/K}\frac{1-e^{-T/t+jn2\pi}}{1-e^{-T/Kt+jn2\pi/K}}\ .$$
If $n$ is an integer (you didn't actually say so) then this is
$$\eqalign{e^{jn\pi/K}\frac{1-e^{-T/t}}{1-e^{-T/Kt+jn2\pi/K}}
  &=e^{T/2Kt}\frac{1-e^{-T/t}}{e^{T/2Kt-jn\pi/K}-e^{-T/2Kt+jn\pi/K}}\cr
  &=e^{T/2Kt}\frac{1-e^{-T/t}}{2\sinh(T/2Kt-jn\pi/K)}\cr
  &=e^{T/2Kt}\frac{1-e^{-T/t}}{2\sinh(T(1-jnwt)/2Kt)}\ .\cr}$$
This is fairly close to what you want, hopefully your constant $A$ will do the rest.  Good luck!

