4. Compute $\sum_{k=1}^{\infty} \frac{k^{4}}{k!}$.
Answer: $15 e$
Solution: Define, for non-negative integers $n$,
$$
S_{n}:=\sum_{k=0}^{\infty} \frac{k^{n}}{k!},
$$
where $0^{0}=1$ when it occurs. Then $S_{0}=e$, and, for $n \geq 1$,
$$
S_{n}=\sum_{k=0}^{\infty} \frac{k^{n}}{k!}=\sum_{k=1}^{\infty} \frac{k^{n}}{k!}=\sum_{k=0}^{\infty} \frac{(k+1)^{n}}{(k+1)!}=\sum_{k=0}^{\infty} \frac{(k+1)^{n-1}}{k!}=\sum_{i=0}^{n-1}\binom{n-1}{i} S_{i},
$$
so we can compute inductively that $S_{1}=e, S_{2}=2 e, S_{3}=5 e$, and $S_{4}=15 e$.