Doubts about this summation (Geometric series)

Question

I don´t understand why does in the numerator is $e^{{t}/{2}}$

Answer 1

$ \sum\limits_{n=1}^{\infty} ar^{n-1}=a\sum\limits_{n=0}^{\infty} r^{n}=a\frac 1 {1-r}$ for $|r|<1$. Now multiply both sides by $r$ to get $ \sum\limits_{n=1}^{\infty} ar^{n}=a\frac r {1-r}$.