CDF from generating function Is there a way to obtain the CDF of a discrete random variable directly from one of its generating functions?

Why, yes! The straightforward, yet not too practical procedure is to invert the generating function to recover the point mass function, and sum it appropriately to obtain the cumulative distribution function.
But suppose you are dealing with a non-negative integer-valued random variable $X$ and have its probability generating function $\mathcal{P}_X(z) = \sum_{n=0}^\infty z^n \Pr(X=n)$ available. Then
$$\begin{eqnarray}
    \frac{\mathcal{P}_X(z)}{1-z} &=& \sum_{n=0}^\infty \sum_{m=0}^\infty z^{n+m} \Pr(X=n) = \sum_{k=0}^\infty \sum_{n=0}^\infty \sum_{m=0}^\infty z^{k} \delta_{k,n+m}\Pr(X=n) \\
   &=& \sum_{k=0}^\infty z^k \sum_{n=0}^k \Pr(X=n) = \sum_{k=0}^\infty z^k \Pr(X\leqslant k) 
\end{eqnarray}
$$
that is the $\mathcal{P}_X(z)/(1-z)$ is the generating function for the cumulative distribution function $F_X(k) = \Pr(X\leqslant k)$:
$$
     F_X(k) = [z^k] \frac{\mathcal{P}_X(z)}{1-z}
$$
For example, consider geometric distribution with probability generating function 
$$
    \mathcal{P}\left(z\right) = \frac{p}{1-(1-p)z}
$$
Hence the cdf is
$$
    [z^k] \frac{p}{\left(1-z\right)\left(1-(1-p)z\right)} = [z^k]\left(\frac{1}{1-z} - \frac{1-p}{1-(1-p)z}\right) = 1-(1-p)^{k+1}
$$
