已知数列｛an｝满足an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(2n) 1.数列｛an｝是递增数

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/04/29 01:52:49

已知数列｛an｝满足an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(2n) 1.数列｛an｝是递增数列还是递减数列
2.证明 an≥1/2对一切正整数恒成立

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,数列是递增数列.
2.
证：
n=1时,a1=1/(1+1)=1/2,不等式成立.
由第1问得数列是递增数列,即n≥2时,an>a1 an>1/2,不等式成立.
综上,得an≥1/2对一切正整数n恒成立.
再问：请问两式子相减之后 1/（2n+1）是哪来的？
再答：由an=1/(n+1)+1/(n+2)+...+1/(2n) 知a(n+1)=1/[(n+1)+1]+1/[(n+1)+2]+...+1/[2(n+1)] a(n+1)=1/[(n+1)+1]+1/[(n+1)+2]+...+1/[2(n+1)] =1/(n+2)+1/(n+3)+...+1/(2n)+1/(2n+1)+1/(2n+2) 比an多了1/(2n+1)和1/(2n+2)两项，少了1/(n+1)一项。

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