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Algebra Difficulty 5.7 AIME, harder Prove it

10. (20 points) Given the sequence {an}\left\{a_{n}\right\} satisfies a1=13,an+1=an+an2n2(n∈N+)a_{1}=\frac{1}{3}, a_{n+1}=a_{n}+\frac{a_{n}^{2}}{n^{2}}\left(n \in \mathbf{N}_{+}\right).
Prove: For all n∈N+n \in \mathbf{N}_{+}, we have
(1) an<12−14na_{n}<\frac{1}{2}-\frac{1}{4 n}.

Solution

10. (1)Obviously,
an>0,an+1=an+an2n2>an. Hence ak+1=ak+ak2k21a1−∑k=1n−11k2>3−[1+∑k=2n−11k(k−1)]=3−[1+∑k=2n−1(1k−1−1k)]=3−(1+1−1n−1)=nn−1>1. \begin{array}{l} a_{n}>0, a_{n+1}=a_{n}+\frac{a_{n}^{2}}{n^{2}}>a_{n} . \\ \text { Hence } a_{k+1}=a_{k}+\frac{a_{k}^{2}}{k^{2}}\frac{1}{a_{1}}-\sum_{k=1}^{n-1} \frac{1}{k^{2}}>3-\left[1+\sum_{k=2}^{n-1} \frac{1}{k(k-1)}\right] \\ =3-\left[1+\sum_{k=2}^{n-1}\left(\frac{1}{k-1}-\frac{1}{k}\right)\right] \\ =3-\left(1+1-\frac{1}{n-1}\right) \\ =\frac{n}{n-1}>1 . \end{array}

Therefore, an14=12−14a_{n}\frac{1}{4}=\frac{1}{2}-\frac{1}{4}.
From $ank2k2+1⇒1ak−1ak+1=akk2ak+1>1k2+1⇒1an=1a1−∑k=1n−1(1ak−1ak+1)n2n+1=12−12(2\$a_{n}\frac{k^{2}}{k^{2}+1} \\ \Rightarrow \frac{1}{a_{k}}-\frac{1}{a_{k+1}}=\frac{a_{k}}{k^{2} a_{k+1}}>\frac{1}{k^{2}+1} \\ \Rightarrow \frac{1}{a_{n}}=\frac{1}{a_{1}}-\sum_{k=1}^{n-1}\left(\frac{1}{a_{k}}-\frac{1}{a_{k+1}}\right) \\ \quad\frac{n}{2 n+1}=\frac{1}{2}-\frac{1}{2(2} n+1)}$
>12−14n.  >\frac{1}{2}-\frac{1}{4 n} \text {. }

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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.