Suppose we want to find

\[\sum \limits_{k=0}^{k=\infty} k e^{-13} \frac{13^k}{k!}\]

Seeing the $13$ in two distinct locations within the summand, we define the generating function $F(z)$ as

\[F(x) = \sum \limits_{k=0}^{\infty} k e^{-x} \frac{x^k}{k!}\]

We first pull out the $e^{-x}$ factor,

\[F(x) = e^{-x} \sum \limits_{k=0}^{\infty} \frac{k x^k}{k!}\]

Looking at the summand now, we see that it resembles the Maclaurin series for $g(x) = e^x$. The only problem is that there is an additional factor of $k$.

\[F(x) = e^{-x } \sum \limits_{k=0}^{\infty} \frac{x^k}{(k-1)!} = x e^{-x} \sum \limits_{k=0}^{\infty} \frac{x^{k-1}}{(k-1)!} = x\]

This means that

\[F(13) = \sum \limits_{k=0}^{k=\infty} k e^{-13} \frac{13^k}{k!} = 13\]

Note that $F(x)$ is basically finding the expected value of a Poisson distribution with Poisson parameter $\lambda = x$. A well-known fact about the Poisson distribution is that its expected value and variance are both equal to the Poisson parameter $\lambda$.