Let $X$ be the outcome of a fair dice roll. Thus, we have $\mathbb{P}(X = i) = 1/6, ~ i = 1,\ldots,6$.
The probability-generating function of $X$ is defined as
\begin{align}
P_X(z) := \mathbb{E}[z^X] = \sum_{x = 1}^6 \mathbb{P}(X = x) z^x = 1/6 \sum_{x = 1}^6 z^x = \frac{1}{6} \frac{z(z^6-1)}{z-1},
\end{align}
with $|z| \le 1$.
The moment-generating function of $X$ is defined as
\begin{align}
M_X(t) := \mathbb{E}[e^{tX}] = \sum_{x = 1}^6 \mathbb{P}(X = x) e^{tx} = 1/6 \sum_{x = 1}^6 e^{tx} = \frac{1}{6} \frac{e^t(e^{6t} - 1)}{e^t - 1},
\end{align}
with $t \in \mathbb{R}$.
The relation between the two generating functions is $P_X(e^t) = \mathbb{E}[e^{tX}] = M_X(t)$. Using the probability-generating function, one can compute the probability mass function and so-called $k$th factorial moments, see the wiki. The moment-generating function allows us to compute the moments of a random variable, see the wiki.
Answering the question in the comment
Computing the sums (in my opinion) is all about knowing a few tricks. I will show you how to do it for the two simplifications of the sum above. We assume that $|a| < 1$.
\begin{align}
I &= \sum_{i = 1}^n a^i = a \sum_{i = 0}^{n-1} a^i = a(1 + a + a^2 + \ldots + a^{n-1}) \\
&= a(1 + a + a^2 + \ldots + a^{n-1} + a^n - a^n ) \\
&= a(1 + I - a^n) = aI + a(1 - a^n) \\
\Rightarrow I(1-a) &= a(1-a^n) \\
\Rightarrow I &= \frac{a(1-a^n)}{(1-a)} = \frac{a(a^n-1)}{a-1}
\end{align}
