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Algebra Difficulty 6.5 National olympiad Prove it

Example 1 Let the sequence a0,a1,a2,,ana_{0}, a_{1}, a_{2}, \cdots, a_{n} satisfy a0=12a_{0}=\frac{1}{2}, and ak+1=ak+a_{k+1}=a_{k}+ 1nak2,k=0,1,2,,n1\frac{1}{n} a_{k}^{2}, k=0,1,2, \cdots, n-1, prove that: 11n<an<11-\frac{1}{n}<a_{n}<1. (1980 Mathematics Olympiad of Finland, UK, Germany, Hungary, and Sweden)

Solution

Proof: By induction, it is easy to see that a01an11n>1an22n>>1a01=1a_{0}\frac{1}{a_{n-1}}-\frac{1}{n}>\frac{1}{a_{n-2}}-\frac{2}{n}>\cdots>\frac{1}{a_{0}}-1=1, i.e., anak(1+ak+1n+1)a_{n}a_{k}\left(1+\frac{a_{k+1}}{n+1}\right). Therefore, we have
1ak+1n+1n+2=11n+2>11n\frac{1}{a_{k+1}}\frac{n+1}{n+2}=1-\frac{1}{n+2}>1-\frac{1}{n}.
In summary, 11n<an<11-\frac{1}{n}<a_{n}<1.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.