# 代数系列——阿贝尔变换的具体运用(1)

## 阿贝尔部分求和

$\sum_{i=1}^{n}{a_{i}b{j}=b_{n}S{n}+\sum_{k=1}^{n-1}{S_n(b_k-b_{k-1})}}$

$a_k=S_k-S_{k-1}$$k=1,2,\cdots,n$得知

\begin{aligned} \sum_{i=1}^{n}a_{i}b_{j}&=\sum_{i=1}^{n}(S_k-S_{k-1})b_{k}\\ &=\sum_{i=1}^{n}S_{k}b_{k}-\sum_{i=1}^{n}S_{k-1}b_{k}\\ &=\sum_{i=1}^{n}S_{k}b_{k}-\sum_{i=1}^{n-1}S_{k}b_{k+1}(这是因为S_0=0,所以可以去掉S_0b_1这项，从而进行指标变换)\\ &=S_nb_n+\sum_{i=1}^{n-1}S_{k}(b_{k}-b_{k-1}) \end{aligned}

## 例题

（1989年全国高中联赛）已知$x_i\in R,i=1,2,\cdots,n,n\geq 2$满足

$\sum_{i=1}^{n}\mid x_{i}\mid=1,\sum_{i=1}^{n}{x_i}=0$

$\mid{\sum_{i=1}^{n}\frac{x_i}{i}} \mid\leq{\frac{1}{2}-\frac{1}{2n}}$

$S_k=\sum_{i=1}^{k}x_i,k=1,2,\cdots,n$，另记$S_0=0$

$|S_k|\leq \frac{1}{2}$

$\sum_{i=1}^{n}\frac{x_i}{i}=\frac{1}{n}S_{n}+\sum_{i=1}^{n-1}(\frac{1}{i}-\frac{1}{i-1})=\sum_{i=1}^{n-1}(\frac{1}{i}-\frac{1}{i-1})$

$|\sum_{i=1}^{n}\frac{x_i}{i}|\leq \sum_{i=1}^{n-1}|S_i|(\frac{1}{i}-\frac{1}{i-1})\leq \frac{1}{2} \sum_{i=1}^{n-1}(\frac{1}{i}-\frac{1}{i-1})=\frac{1}{2}-\frac{1}{2n}$

## Conclusion

posted @ 2021-03-13 15:04  SaikyoDOROC  阅读(52)  评论(0编辑  收藏