数学归纳法证不等式1/(n+1)+1/(n+2)+...+1/(3n+1)>1

逸枫521 1年前他留下的回答 已收到1个回答

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1.当n=1时,1/2 + 1/3 + 1/4 = 13/12 > 1,所以当n=1时不等式成立
2.假设当n=k-1时不等式成立,
即1/[(k-1)+1]+1/[(k-1)+2]+...+1/[3(k-1)+1]=1/k+1/(k+1)+...+1/(3k-2)>1
则当n=k时,
1/(k+1)+1/(k+2)+...+1/(3k-1)+1/3k+1/(3k+1)
=[1/k+1/(k+1)+1/(k+2)+...+1/(3k-2)]+1/(3k-1)+1/3k+1/(3k+1)-1/k
>1+1/(3k-1)+1/3k+1/(3k+1)-1/k
又因为
1/(3k-1)+1/3k+1/(3k+1)-1/k（通分）=2/[3k(3k+1)(3k-1)]>0
所以
1/(k+1)+1/(k+2)+...+1/(3k-1)+1/3k+1/(3k+1)>0
即当n=k时,不等式也成立
综上所诉,不等式1/(n+1)+1/(n+2)+...+1/(3n+1)>1 成立
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终于打完了,希望能看懂

1年前他留下的回答

5

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