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Romanian Mathematical Olympiad - Final Round

Algebra Difficulty 5.1 AIME, harder Prove it Romania

a) Prove that $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{2^m} < m$ , for all $m \in \mathbb{N}^*$ .

b) Let $p_1, p_2, \dots, p_n$ be the sequence of the primes less than $2^{100}$ . Prove that
$\frac{1}{p_1} + \frac{1}{p_2} + \dots + \frac{1}{p_n} < 10.$

Solution

a) It is a well-known inequality that can be immediately proved by induction.

b) The numbers $p_i p_j p_k p_l$ , with $1 \le i \le j \le k \le l \le n$ , are distinct and all less than $2^{400}$ , so
$\left(\frac{1}{p_1} + \frac{1}{p_2} + \dots + \frac{1}{p_n}\right)^4 \le 4! \sum_{1 \le i \le j \le k \le l \le n} \frac{1}{p_i p_j p_k p_l} < 24 \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2^{400}}\right).$
The proof finishes noticing that $24 \left(\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{2^{400}}\right) < 24 \cdot 400 < 10000.$

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