阿贝尔分布求和公式
2018.04.10
- 设\(s_{k}=a_{1}+a_{2}+\cdots+a_{k},k=1,2,\cdots,n\),则\(\sum_{k=1}^{n}a_{k}b_{k}=s_{n}b_{n}+\sum_{k=1}^{n-1}s_{k}(b_{k}-b_{k+1})\).
- 设\(s_{n}=a_{1}+a_{2}+\cdots+a_{n}\to s (n\to \infty)\),则\(\sum_{k=1}^{n}a_{k}b_{k}=sb_{1}+(s_{n}-s)b_{n}-\sum_{k=1}^{n-1}(s_{k}-s)(b_{k+1}-b_{k})\)
- 设\(\varphi(n)>0,\varphi(n)\uparrow \infty(n \to \infty)\),且\(\sum_{n=1}^{\infty}a_{n}\),则\(\sum_{k=1}^{n}a_{k}\varphi(k)=o(\varphi(n)),(n\to \infty)\).
- 设\(\varphi(n)\downarrow 0(n \to \infty)\),且\(\sum_{n=1}^{\infty}a_{n}\varphi(n)\)收敛,则\(\lim_{n\to\infty}(a_{1}+a_{2}+\cdots+a_{n})\varphi(n)=0\).
第三个命题的证明要“分两步走”
更多相关命题可参考阿贝尔分布求和法的应用1