Proof of this (geometric) infinite sum 
Question: Prove that when $\exp(\Re[st])>1$:
$$\sum_{n=0}^{\infty}\exp\left[-nts\right]=\frac{e^{st}}{e^{st}-1}$$

I think that it is an geomeric sum, but I new to this kind of finding a sum. Thanks for any help.
 A: It's exactly a geometric sum:
$$
\sum_{n=0}^\infty e^{-nts}=\sum_{n=0}^\infty (e^{-ts})^n=\frac1{1-e^{-ts}}=\frac{e^{ts}}{e^{ts}+1}.
$$
For the series to converge and the above to work you need $|e^{-ts}|<1$. This is $|e^{ts}|>1$. Since 
$$
|e^{ts}|=|e^{\text{Re}\,ts+i\,\text{Im}\,ts}|=|e^{\text{Re}\,ts}|>1,
$$
the  hypothesis guarantees that the series converges. 
A: Observe that if $\;|z|<1\;$ then
$$\sum_{n=0}^\infty z^n=\frac1{1-z}$$
In our case,
$$e^{\text{Re}\,(st)}>1\implies \left|e^{-st}\right|=e^{-\text{Re}\,(st)}<1\implies$$
$$\sum_{n=0}^\infty\left(e^{-st}\right)^n=\frac1{1-e^{-st}}=\frac{e^{st}}{e^{st}-1}$$
A: The theorem you want to use (the geometric series) is: 
$$\sum_{n=0}^{\infty} z^n = \frac{1}{1-z},$$
for $z \in \mathbb{C}$ such that $|z|<1$. Note that 
$$e^{-nts}=(e^{-ts})^n$$
for $n \in \mathbb{N}$. Substituting $z=e^{-ts}$ we obtain the required result, if of course $|e^{-st}|<1$. To see this let $st = x+iy$. We are given that $x>1$. So $$|e^{-st}| = |e^{-x-iy}|=e^{-x}<1.$$
