给定整数 $n>2$，设正实数 $a_1,a_2,\cdots,a_n$ 满足 $a_k \leqslant 1$，$k=1,2,\cdots,n$，记 $A_k=\dfrac {a_1+a_2+\cdots+a_k}{k}$，$k=1,2,\cdots,n$．求证：$\displaystyle \left|\sum \limits_{k=1}^{n}a_k-\sum \limits_{k=1}^{n}A_k \right|<\dfrac {n-1}{2}$．
解析 根据题意，对 $k=1,2,\cdots,n-1$，有\[|A_n-A_k|=\left|\left(\dfrac 1n-\dfrac 1k \right)\sum \limits_{i=1}^{k}a_i+\dfrac 1n\sum \limits_{i=k+1}^{n}a_i \right|=\left|\dfrac 1n\sum \limits_{i=k+1}^{n}a_i-\left(\dfrac 1k-\dfrac 1n \right)\sum \limits_{i=1}^{k}a_i \right|,\]而\[\begin{cases} 0<\dfrac 1n\displaystyle\sum_{i=k+1}^na_i\leqslant\dfrac{n-k}n,\\ 0<\left(\dfrac 1k-\dfrac 1n\right)\displaystyle\sum_{i=1}^ka_i\leqslant \left(\dfrac 1k-\dfrac 1n\right)k=\dfrac{n-k}n,\end{cases}\]因此\[|A_n-A_k|<\dfrac{n-k}n,\] 故\[\left|\sum \limits_{k=1}^{n}a_k-\sum \limits_{k=1}^{n}A_k \right|=\left|nA_n-\sum \limits_{k=1}^{n}A_k \right|=\left| \sum \limits_{k=1}^{n-1}(A_n-A_k) \right|\leqslant \sum \limits_{k=1}^{n-1} \left|A_n-A_k \right|< \sum \limits_{k=1}^{n-1}\dfrac{n-k}n=\dfrac {n-1}{2}.\]