求证1/(n+1)+1/(n+2)+...+1/(3n+1)>1

n=1时
1/2+1/3+1/4>1/2+1/4+1/4=1,原命题成立
假设当n=k时原命题成立
当n=k+1时
1/(k+1+1)+1/(k+1+2)+...+1/(3k+4)
=1/(k+1)+1/(k+2)+...+1/(3k+1)+1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)
＞1+1/(3k+2)+1/(3k+3)+1/(3k+4)-1/(k+1)
证明:1/(3k+2)+1/(3k+3)+1/(3k+4)>1/(k+1)
<=> 1/(3k+2)-1/(3k+3)>1/(3k+3)-1/(3k+4)
<=>1/(3k+2)(3k+3)>1/(3k+3)(3k+4)
上式显然成立
故当n=k+1时也成立
故对于所以的n(n为正整数)均成立