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Algebra Difficulty 5.6 AIME, harder Prove it

Example 12 Let an=∑k=1n1k(n+1−k)a_{n}=\sum_{k=1}^{n} \frac{1}{k(n+1-k)}. Prove that for any positive integer n⩾2n \geqslant 2, an+1<ana_{n+1}<a_{n}.

Solution

Note that 1k(n+1−k)=1n+1(1k+1n+1−k)\frac{1}{k(n+1-k)}=\frac{1}{n+1}\left(\frac{1}{k}+\frac{1}{n+1-k}\right),
thus, an=2n+1∑k=1n1ka_{n}=\frac{2}{n+1} \sum_{k=1}^{n} \frac{1}{k}.
Therefore, for any positive integer n⩾2n \geqslant 2, we have 12(an−an+1)=1n+1∑k=1n1k−1n+2∑k=1n+11k=\frac{1}{2}\left(a_{n}-a_{n+1}\right)=\frac{1}{n+1} \sum_{k=1}^{n} \frac{1}{k}-\frac{1}{n+2} \sum_{k=1}^{n+1} \frac{1}{k}= (1n+1−1n+2)∑k=1n1k−1(n+1)(n+2)=1(n+1)(n+2)(∑k=1n1k−1)>0\left(\frac{1}{n+1}-\frac{1}{n+2}\right) \sum_{k=1}^{n} \frac{1}{k}-\frac{1}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}\left(\sum_{k=1}^{n} \frac{1}{k}-1\right)>0,
so an+1<ana_{n+1}<a_{n}.

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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.