Question

设an=1/（n+1）+1/（n+2)+1/(n+3)+...+1/2n(n属于正整数）则an+1-an=--

Answer 1

1.
a(n+1)-an
=1/[(n+1)+1]+1/[(n+1)+2]+...+1/[(n+1)+(n-1)]+1/[(n+1)+n]+1/[(n+1)+(n+1)]
-[1/(n+1)+1/(n+2)+...+1/(2n)]
=1/(n+2)+1/(n+3)+...+1/(2n)+1/(2n+1)+1/(2n+2)-[1/(n+1)+1/(n+2)+...+1/(2n)]
=1/(2n+1)+1/(2n+2)
-1/(n+1)
=1/(2n+1)-1/(2n+2)>0
a(n+1)>an
数列
递增数列
2.
证：
n=1
a1=1/(1+1)=1/2
等式
立
由第1问
数列
递增数列
即n≥2
an>a1
an>1/2
等式
立
综
an≥1/2
切
整数n恒
立