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RO-TST 1996 Problem 8

    Let p1,p2,…,pkp_1,p_2,\ldots,p_k be the distinct prime divisors of nn and let an=1p1+1p2+⋯+1pka_n=\frac {1}{p_1}+\frac {1}{p_2}+\cdots+\frac {1}{p_k} for n≥2n\geq 2 . Show that for every positive integer N≥2N\geq 2 the following inequality holds: ∑k=2Na2a3⋯ak<1\sum_{k=2}^{N} a_2a_3 \cdots a_k <1

    Laurentiu Panaitopol

    Dataset text AI-MO AoPS. Quoted with source.

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