Sum of exponentials - derivation

Question

In a couple of scientific papers, I've found this sum identity involving exponentials:

$\sum_{n=0}^{\frac{E_{BD}}{\Delta E}} e^{\gamma(n \Delta E - E_{BD})} = \frac{1}{1 - e^{\gamma\Delta E}}$

Screenshot

However, I can't find its derivation anywhere and I'm not able to work it out on my own.

Could someone please explain it to me? :)

Thanks in advance!

Answer 1

I don't know anything about physics, but evaluating the sum

$$ \sum_{n=0}^{\frac{E_{BD}}{\Delta E}} e^{\gamma \left(n\Delta E - E_{BD}\right)} = e^{-\gamma E_{BD}} \sum_{n=0}^{\frac{E_{BD}}{\Delta E}} \left(e^{\gamma \Delta E}\right)^{n} = e^{-\gamma E_{BD}} \frac{1-e^{\gamma \Delta E\left(\frac{E_{BD}}{\Delta E}+1\right)}}{1-e^{\gamma \Delta E}} = e^{-\gamma E_{BD}}\frac{1-e^{\gamma E_{BD} + \gamma \Delta E}}{1-e^{\gamma \Delta E}} = \frac{e^{-\gamma E_{BD}} - e^{\gamma\Delta E}}{1-e^{\gamma \Delta E}} $$ Then you would have to argue that the numerator is equal to one.