用数学归纳法证明证明1/(n+1)+1/(n+2)+.+1/(3n+1)≥1（n∈N）

Question

用数学归纳法证明
证明1/(n+1)+1/(n+2)+.+1/(3n+1)≥1 （n∈N）

Answer 1

证明：
当n=1时,1/2 + 1/3 +1/4=13/12＞1,结论成立.
令An=1/(n+1)+1/(n+2)+.+1/(3n+1)
假设当n=k时结论成立,即
Ak=1/(k+1)+1/(k+2)+…+1/(3k+1)>1
我们来证明n=k+1时,结论也成立
因为
A(k+1)=1/(k+2)+1/(k+3)+…+1/(3k+4)
=[1/(k+1)+1/(k+2)+…+1/(3k+1)]+1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)
=Ak +1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)
下面我们来证明1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)＞0 ①
①式可化左端可化为
1/(3k+3-1)+1/(3k+3)+1/(3k+3+1)-3/(3k+3)
=1/(3k+3-1)+1/(3k+3+1)-2/(3k+3) ②
令a=3k+3
若1/（a-1) +1/(a+1)＞2/a (其中a＞1) 成立
则②>0
1/（a-1) +1/(a+1)=2a/(a²-1)＞2a/a²=2/a
这样1/（a-1) +1/(a+1)＞2/a成立,从而②式大于0,即①式成立,从而
A(k+1)＞Ak＞1