infinite summation of exponential functions

Question

I've discovered through Wolfram Alpha that

$\sum_{t=1}^{\infty}{e^{-bt}}=\frac{1}{e^b-1}$

What are the steps of derivation here? According to infinite summation of power series:

$\sum_{t=1}^{\infty}p^t=\frac{1}{1-p}$,

I expected the solution to be

$\sum_{t=1}^{\infty}{(e^{-b})^t}=\frac{1}{1-e^{-b}}$.

What am I getting wrong?

In extension, how do I derive

$\sum_{t=1}^{\infty}{e^{-b(t-1)}}$ ?

Answer 1

Your second formula isn't quite right: if $|p|<1$, then $$ \sum_{t=1}^{\infty}p^t=\frac{p}{1-p}$$ Using this with $p=e^{-b}$ yields $$ \sum_{t=1}^{\infty}e^{-bt}=\frac{e^{-b}}{1-e^{-b}}=\frac{1}{e^b-1}$$ as claimed.