China Mathematical Competition (Complementary Test)

For $1\le k\le n-1$, we have $0<{\sum }_{i=1}^k{a}_i\le k$ and $0<{\sum }_{i=k+1}^n{a}_i\le n-k$. By using the fact that $|x-y|<\max \{x,y\}$ for $x,y>0$, we get $$\begin{array}{rl}|A_n-A_k|& =\left|\left(\frac{1}{n}-\frac{1}{k}\right)\sum _{i=1}^k{a}_i+\frac{1}{n}\sum _{i=k+1}^n{a}_i\right|\\ & =\left|\frac{1}{n}\sum _{i=k+1}^n{a}_i-\left(\frac{1}{k}-\frac{1}{n}\right)\sum _{i=1}^k{a}_i\right|\\ & <\max \left\{\frac{1}{n}\sum _{i=k+1}^n{a}_i,\left(\frac{1}{k}-\frac{1}{n}\right)\sum _{i=1}^k{a}_i\right\}\\ & \le \max \left\{\frac{1}{n}(n-k),\left(\frac{1}{k}-\frac{1}{n}\right)k\right\}\\ & =1-\frac{k}{n}.\end{array}$$ Therefore, $$\begin{array}{rl}\left|\sum _{k=1}^n{a}_k-\sum _{k=1}^n{A}_k\right|& =\left|n{A}_n-\sum _{k=1}^n{A}_k\right|\\ & =\left|\sum _{k=1}^{n-1}(A_n-A_k)\right|\le \sum _{k=1}^{n-1}\left|A_n-A_k\right|\\ & <\sum _{k=1}^{n-1}\left(1-\frac{k}{n}\right)=\frac{n-1}{2}.\end{array}$$ This completes the proof. ☐