IMC 2002 · Problem 3

Day 29th IMC · Warsaw, Poland

Statement

For each n≥1n \ge 1 let

an=∑k=0∞knk!,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 an⋅bna_n \cdot b_n is an integer.

Official solution

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