(1/2+1/3+...+1/2008)(1+1/2+1/3+...1/2007)与(1+1/2+1/3+...+1/2008)(1/2+1/3+...+1/2007)的大小

问题描述:

(1/2+1/3+...+1/2008)(1+1/2+1/3+...1/2007)与(1+1/2+1/3+...+1/2008)(1/2+1/3+...+1/2007)的大小

令a=1/2+1/3+...+1/2007
则(1/2+1/3+...+1/2008)(1+1/2+1/3+...1/2007)-(1+1/2+1/3+...+1/2008)(1/2+1/3+...+1/2007)
=(a+1/2008)(1+a)-(1+a+1/2008)*a
=a(a+1)+(a+1)/2008-a(a+1)-a/2008
=(a+1)/2008-a/2008
=(a+1-a)/2008
=1/2008>0
所以(1/2+1/3+...+1/2008)(1+1/2+1/3+...1/2007)>(1+1/2+1/3+...+1/2008)(1/2+1/3+...+1/2007)