| 两侧同时换到之前的修订记录前一修订版 | |
| notes:math:199_200vs200_199 [2021/06/02 06:08] – delphij | notes:math:199_200vs200_199 [2022/01/15 01:28] (当前版本) – delphij |
|---|
| <m>r = (1+1/2)*(1+1/3)*(1+1/4)*...*(1+1/199)*(1/199)</m> | <m>r = (1+1/2)*(1+1/3)*(1+1/4)*...*(1+1/199)*(1/199)</m> |
| |
| 或换言之将前198项替换为通项公式为 <m>1+1/(n+1)<m> 或 <m>(n+2)/(n+1)</m>，由于前198项每项都大于<m>1+1/199=200/199</m>，若能证明<m>r<1</m>，就可以证明<m>q<1</m>了。我们有： | 或换言之将前198项替换为通项公式为 <m>1+1/(n+1)</m> 或 <m>(n+2)/(n+1)</m>，由于前198项每项都大于<m>1+1/199=200/199</m>，若能证明<m>r<1</m>，就可以证明<m>q<1</m>了。我们有： |
| |
| <m>r = (1+1/2)*(1+1/3)*(1+1/4)*...*(1+1/198)*(1+1/199)*(1/199) = (3/2)*(4/3)*(5/4)*...*(199/198)*(200/199)*(1/199)</m> | <m>r = (1+1/2)*(1+1/3)*(1+1/4)*...*(1+1/198)*(1+1/199)*(1/199) = (3/2)*(4/3)*(5/4)*...*(199/198)*(200/199)*(1/199)</m> |