比较审敛法










无穷级数

ζ(s)=∑k=1∞1ks{displaystyle zeta (s)=sum _{k=1}^{infty }{frac {1}{k^{s}}}}zeta (s)=sum _{k=1}^{infty }{frac {1}{k^{s}}}


无穷级数










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比较审敛法是一种判定级数是否收敛的方法。



定理


设两个正项级数∑n=1∞un{displaystyle sum _{n=1}^{infty }u_{n}}sum _{n=1}^{infty }u_{n}和∑n=1∞vn{displaystyle sum _{n=1}^{infty }v_{n}}{displaystyle sum _{n=1}^{infty }v_{n}}，且un≤vn(n=1,2,3,...){displaystyle u_{n}leq v_{n}(n=1,2,3,...)}{displaystyle u_{n}leq v_{n}(n=1,2,3,...)}：


如果级数∑n=1∞vn{displaystyle sum _{n=1}^{infty }v_{n}}{displaystyle sum _{n=1}^{infty }v_{n}}收敛，则级数∑n=1∞un{displaystyle sum _{n=1}^{infty }u_{n}}sum _{n=1}^{infty }u_{n}收敛；


如果级数∑n=1∞un{displaystyle sum _{n=1}^{infty }u_{n}}sum _{n=1}^{infty }u_{n}发散，则级数∑n=1∞vn{displaystyle sum _{n=1}^{infty }v_{n}}{displaystyle sum _{n=1}^{infty }v_{n}}发散。



证明


设σk=∑n=1∞un,sk=∑n=1∞vn{displaystyle sigma _{k}=sum _{n=1}^{infty }u_{n},s_{k}=sum _{n=1}^{infty }v_{n}}{displaystyle sigma _{k}=sum _{n=1}^{infty }u_{n},s_{k}=sum _{n=1}^{infty }v_{n}}当un≤vn{displaystyle u_{n}leq v_{n}}{displaystyle u_{n}leq v_{n}}时，则有σk≤sk{displaystyle sigma _{k}leq s_{k}}{displaystyle sigma _{k}leq s_{k}}：


当级数∑n=1∞vn{displaystyle sum _{n=1}^{infty }v_{n}}{displaystyle sum _{n=1}^{infty }v_{n}}收敛时，数列sk{displaystyle s_{k}}{displaystyle s_{k}}有界，从而数列σk{displaystyle sigma _{k}}sigma _{k}有界，所以级数∑n=1∞un{displaystyle sum _{n=1}^{infty }u_{n}}sum _{n=1}^{infty }u_{n}收敛；


当级数∑n=1∞un{displaystyle sum _{n=1}^{infty }u_{n}}sum _{n=1}^{infty }u_{n}发散时，数列σk{displaystyle sigma _{k}}{displaystyle sigma _{k}}无界，从而数列sk{displaystyle s_{k}}{displaystyle s_{k}}无界，所以级数∑n=1∞vn{displaystyle sum _{n=1}^{infty }v_{n}}{displaystyle sum _{n=1}^{infty }v_{n}}发散。



参见



  • 审敛法

  • 比值审敛法

  • 根值审敛法

  • 交错级数审敛法




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