证明：1/(x+1)+1(x+2)…+1/(3n+1)>=1证明：当n=1时,1/2+1/3+1/4=13/12＞1,结论成立.假设当n=k时结论成立,即Sk=1/(k+1)+1/(k+2)+…+1/(3k+1)>1我们来证明n=k+1时,结论也成立（我们会证明S(k+1)＞Sk）因为S(k+1)=1/(k+2

Question

证明： 1/(x+1)+1(x+2)…+1/(3n+1)>=1
证明：
当n=1时,1/2 + 1/3 +1/4=13/12＞1,结论成立.
假设当n=k时结论成立,即
Sk=1/(k+1)+1/(k+2)+…+1/(3k+1)>1
我们来证明n=k+1时,结论也成立（我们会证明S(k+1)＞Sk）
因为
S(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)
=Sk +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
为了要证明②式大于0,我们只需证明更一般的
1/（a-1) +1/(a+1)＞2/a (其中a＞1) ③
(希望读者可看出它与②式的一致）
我们知道1/（a-1) +1/(a+1)=2a/(a²-1)＞2a/a²=2/a
这样③式成立,从而②式大于0,即①式成立,从而
S(k+1)＞Sk＞1 证完.
证明过程中1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)
是怎么得来的啊

Answer 1

题目中不是算了吗?
S(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)
=Sk +1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)
S（k+1）=1/(k+2)+1/(k+3)+…+1/(3k+4)
S（k）=1/(k+1)+1/(k+2)+…+1/(3k+1)
S（k+1）-S（k）=1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)