摘要
本文给出任意项级数收敛判定方法:如果级数∑n=1∞ an的项添加括号后所成的级数收敛且limn→∞an=0,则该级数收敛.由此获得:设C={ai|ai∈Z,i=0,1,…,k},D={a2j|a2j=2r2j+1∈C,r2j∈Z},E={a2j+1|a2j+1=2r2j+1+1∈C,r2j+1∈Z}且|D|=2p+1,|E|=2q,p,q∈Z,则级数∑n=1∞sinπ/2(a0nk+a1nk-1+…+ak)/n发散,否则收敛.同时得到:∑n=1∞sinπ/2n2s+1/n收敛,级数∑n=1∞sinπ/2n2s/n发散,其中s∈N.
A convergence determination method of any term series is presented:If the series ∑n=1∞ an converges after parentheses are added to the terms of the series,and limn→∞an=0,the series converges.The resulting:if C={ai|ai∈Z,i=0,1,…,k},D={a2j|a2j=2r2j+1∈C,r2j∈Z},E={a2j+1|a2j+1=2r2j+1+1∈C,r2j+1∈Z},and |D|=2 p+1,|E|=2 q,p,q∈Z,then the series ∑n=1∞sinπ/2(a0nk+a1nk-1+…+ak)/n divergences,or else convergences.Meanwhile resulting:∑n=1∞sinπ/2n2s+1/n convergences,the series ∑n=1∞sinπ/2 n2 s/n divergences,therein s ∈ N.
作者
杜先云
任秋道
DU Xianyun;REN Qiudao(College of Applied Mathematics, Chengdu University of Information Technology, Chengdu, Sichuan 610225;School of Mathematics and Physics, Mianyang Teachers' College,Mianyang, Sichuan 621000)
出处
《绵阳师范学院学报》
2020年第5期1-4,15,共5页
Journal of Mianyang Teachers＇ College
基金
四川省教育厅基金资助(16ZB0314).
关键词
数列
数列收敛
级数收敛
Sequence
sequence Convergence
Series convergence
