IMC 2002 · Day 2 · P9

We prove by induction on $n$ that $a_n / e$ and $b_n e$ are integers, we prove this for $n = 0$ as well. (For $n = 0$, the term $0^0$ in the definition of the sequences must be replaced by 1.)

From the power series of $e^x$, $a_0 = e^1 = e$ and $b_0 = e^{-1} = 1/e$.

Suppose that for some $n \ge 0$, $a_0, a_1, \dots, a_n$ and $b_0, b_1, \dots, b_n$ are all multipliers of $e$ and $1/e$, respectively. Then, by the binomial theorem, $$a_{n+1} = \sum_{k=0}^{\infty} \frac{(k+1)^{n+1}}{(k+1)!} = \sum_{k=0}^{\infty} \frac{(k+1)^n}{k!} = \sum_{k=0}^{\infty} \sum_{m=0}^{n} \binom{n}{m} \frac{k^m}{k!} = \sum_{m=0}^{n} \binom{n}{m} \sum_{k=0}^{\infty} \frac{k^m}{k!} = \sum_{m=0}^{n} \binom{n}{m} a_m$$ and similarly $$\begin{align*} b_{n+1} &= \sum_{k=0}^{\infty} (-1)^{k+1} \frac{(k+1)^{n+1}}{(k+1)!} = -\sum_{k=0}^{\infty} (-1)^k \frac{(k+1)^n}{k!} = \\ &= -\sum_{k=0}^{\infty} (-1)^k \sum_{m=0}^{n} \binom{n}{m} \frac{k^m}{k!} = -\sum_{m=0}^{n} \binom{n}{m} \sum_{k=0}^{\infty} (-1)^k \frac{k^m}{k!} = -\sum_{m=0}^{n} \binom{n}{m} b_m. \end{align*}$$ The numbers $a_{n+1}$ and $b_{n+1}$ are expressed as linear combinations of the previous elements with integer coefficients which finishes the proof.