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Algebra Difficulty 6.3 National olympiad Prove it

15. {an}\left\{a_{n}\right\} satisfies an+1=12an2−an+2(n⩾1)a_{n+1}=\frac{1}{2} a_{n}^{2}-a_{n}+2(n \geqslant 1). Prove:
(1) If a1=4a_{1}=4, then an+1⩾(32)n⋅ana_{n+1} \geqslant\left(\frac{3}{2}\right)^{n} \cdot a_{n};
(2) If a1=1a_{1}=1, then when n⩾5n \geqslant 5, ∑k=1n1ak<n−1\sum_{k=1}^{n} \frac{1}{a_{k}}<n-1.

Solution

15. (1) First prove an+1⩾ana_{n+1} \geqslant a_{n}, then prove that for n⩾2n \geqslant 2, an+1>2ana_{n+1}>2 a_{n}, and then use mathematical induction to prove 12an>(32)n+1\frac{1}{2} a_{n}>\left(\frac{3}{2}\right)^{n}+1. Therefore, an+1an=12an+2an−1>an2−1>\frac{a_{n+1}}{a_{n}}=\frac{1}{2} a_{n}+\frac{2}{a_{n}}-1>\frac{a_{n}}{2}-1> (32)n\left(\frac{3}{2}\right)^{n}
(2) Prove 1an=1an−2−1an+1−2\frac{1}{a_{n}}=\frac{1}{a_{n}-2}-\frac{1}{a_{n+1}-2}, and then use mathematical induction to prove that for n⩾5n \geqslant 5, an<2−1n−1a_{n}<2-\frac{1}{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.