A Reciprocal Product Recursion

★ 8/10sequences AM-GM Telescoping inequality

Problem

A sequence $\{a_n\}$ satisfies $a_1 = 1$ and

a_{n+1}a_n = \frac{1}{n} \qquad (n \in \mathbb{N}^*).

1. Prove that $\dfrac{a_{n+2}}{n} = \dfrac{a_n}{n+1}$ . 2. Prove that

2\left(\sqrt{n+1} - 1\right) \leqslant \frac{1}{2a_3} + \frac{1}{3a_4} + \cdots + \frac{1}{(n+1)a_{n+2}} \leqslant n.

Answer

Solution

Difficulty	8/10
Topics	sequences, AM-GM, Telescoping, inequality

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