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Algebra Difficulty 4.2 AIME Prove it Soviet Union

Problem:
The sequence {an}\{a_n\} is defined as follows: a1=1a_1 = 1, an+1=an+1/ana_{n+1} = a_n + 1 / a_n for n1n \geq 1. Prove that a100>14a_{100} > 14.

Solution

Solution:
First we must notice that for 1a,b1 \leq a, b we have a<ba < b, then a+1/a<b+1/ba + 1 / a < b + 1 / b. This is basic to any estimation.

The obvious approach is to notice that if aina_i \leq n, then ai+1ai+1/na_{i+1} \geq a_i + 1 / n. Hence it takes at most nn steps to get from n1n-1 to nn. Unfortunately, this does not quite work: we need 2+3++14=1042 + 3 + \ldots + 14 = 104 steps to get from 11 to 1414.

The trick is to notice that an+12>an2+2a_{n+1}^2 > a_n^2 + 2. But a2=2a_2 = 2, so an2>2na_n^2 > 2n. That gives a1002>200>142a_{100}^2 > 200 > 14^2.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.