an=1/(2n-1),求n大于等于2且为正整数时a（n+1）+a（n+2）+……a（2n）的最大值

记S(k)=a(k+1)+a(k+2)+...+a(2k)
S(k+1)=a(k+2)+a(k+3)+...+a(2k+2)
S(k+1)-S(k)
=a(2k+2)+a(2k+1)-a(k+1)
=1/(4k+3)+1/(4k+1)-1/(2k+1)
=(4k+3+4k+1)/(16k^2+16k+3)-1/(2k+1)
=(8k+4)/[(4k+2)^2-1]-1/(2k+1)
>(8k+4)/(4k+2)^2-1/(2k+1)
=1/(2k+1)-1/(2k+1)
=0
可见S(k)随k增加而增加
无最大值
最小值S(2)=a(3)+a(4)=1/5+1/7=12/35