IMC 2002 · Problem 3

Day 29th IMC · Warsaw, Poland

Statement

For each n1n \ge 1 let

an=k=0knk!,bn=k=0(1)kknk!.a_n = \sum_{k=0}^{\infty} \frac{k^n}{k!}, \quad b_n = \sum_{k=0}^{\infty} \frac{(-1)^k k^n}{k!}.

Show that anbna_n \cdot b_n is an integer.

Official solution

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