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 $$\begin{array}{rl}\frac{1}{2}(a_n-a_{n+1})& =\frac{1}{n+1}\sum _{k=1}^n\frac{1}{k}-\frac{1}{n+2}\sum _{k=1}^{n+1}\frac{1}{k}\\ & =\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\sum _{k=1}^n\frac{1}{k}-\frac{1}{(n+1)(n+2)}\\ & =\frac{1}{(n+1)(n+2)}\left(\sum _{k=1}^n\frac{1}{k}-1\right)\\ & >0.\end{array}$$ That means $a_{n+1}<a_n$.