求证:1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24(n是正整数)

求证:1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24(n是正整数)

题目
求证:1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24(n是正整数)
答案
证明:
当k=1时
1/2+1/3+1/4=13/12=26/24>25/24
结论成立.
假设k=n时结论成立,即
1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24
当k=n+1时
由于
9(n+1)^2=9n^2+18n+9>9n^2+18n+8=(3n+2)(3n+4)

9(n+1)^2/[(3n+2)(3n+4)]-1>0
左侧为
1/[(n+1)+1]+1/[(n+1)+2]+1/[(n+1)+3]+...+1/[3(n+1)+1]
=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+{1/(3n+2)+1/(3n+3)+1/(3n+4)-1/(n+1)}
=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+{6(n+1)/[(3n+2)(3n+4)]-2/(3n+3)}
=1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+2/(3n+3)*{9(n+1)^2/[(3n+2)(3n+4)]-1}
>1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24.
结论成立.
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