Question

一道数学题证明谢谢啊

证明
(1+n/2)≤1+1/2+1/3+……+1/2^n≤（1/2+n）

要过程谢谢

Answer 1

n=1时
1+1/2<=1+1/2<=1/2+1
设n=k时
(1+n/2)≤1+1/2+1/3+……+1/2^n≤（1/2+n）
n=k+ 1时
1+(k+1)/2=1/2+(1+k/2)<=1/2+1+1/2+1/3+……+1/2^k+1/2
又1/(2^k+1)+1/(2^k+2)+...+1/2^(k+1)>2^k/2^(k+1)=1/2
1/(2^k+1)+1/(2^k+2)+...+1/2^(k+1)<2^k/2^k=1
故1+(k+1)/2=(1+k/2)+1/2<=1+1/2+1/3+……+1/2^k+1/2
<1/2+1+1/2+1/3+……+1/2^k+1/(2^k+1)+1/(2^k+2)+...+1/2^(k+1)<1+1/2+1/3+……+1/2^k+1<=(1/2+k)+1=1/2+(k+1)
从而1+(k+1)/2≤1+1/2+1/3+……+1/2^(k+1)≤1/2+(k+1)
由数学归纳法知对于所有n>=1,有
(1+n/2)≤1+1/2+1/3+……+1/2^n≤（1/2+n）

Answer 2

呃。。。
几年级啊。。
俺6年级。。