China Mathematical Competition

As
$$\frac{1}{k(n+1-k)} = \frac{1}{n+1}\left(\frac{1}{k} + \frac{1}{n+1-k}\right),$$
we get $a_n = \frac{2}{n+1} \sum_{k=1}^{n} \frac{1}{k}$. Then for $n \ge 2$ we have
$$\begin{aligned} \frac{1}{2}(a_n - a_{n+1}) &= \frac{1}{n+1} \sum_{k=1}^{n} \frac{1}{k} - \frac{1}{n+2} \sum_{k=1}^{n+1} \frac{1}{k} \\ &= \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \sum_{k=1}^{n} \frac{1}{k} - \frac{1}{(n+1)(n+2)} \\ &= \frac{1}{(n+1)(n+2)} \left(\sum_{k=1}^{n} \frac{1}{k} - 1\right) \\ &> 0. \end{aligned}$$
That means $a_{n+1} < a_n$.