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Algebra Difficulty 4.7 AIME Prove it China

Let an=∑k=1n1k(n+1−k)a_n = \sum_{k=1}^{n} \frac{1}{k(n+1-k)}. Prove that an+1<ana_{n+1} < a_n for n≥2n \ge 2.

Solution

As
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),
we get an=2n+1∑k=1n1ka_n = \frac{2}{n+1} \sum_{k=1}^{n} \frac{1}{k}. Then for n≥2n \ge 2 we have
12(an−an+1)=1n+1∑k=1n1k−1n+2∑k=1n+11k=(1n+1−1n+2)∑k=1n1k−1(n+1)(n+2)=1(n+1)(n+2)(∑k=1n1k−1)>0. \begin{aligned} \frac{1}{2}(a_n - a_{n+1}) &= \frac{1}{n+1} \sum_{k=1}^{n} \frac{1}{k} - \frac{1}{n+2} \sum_{k=1}^{n+1} \frac{1}{k} \\ &= \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. \end{aligned}
That means an+1<ana_{n+1} < a_n.

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