用数学归纳法证明1/n+1/(n+1)+1/(n+2)+……1/n²>1(n∈N*,n>1)

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用数学归纳法证明1/n+1/(n+1)+1/(n+2)+……1/n²>1(n∈N*,n>1)用数学归纳法证明1/n+1/(n+1)+1/(n+2)+……1/n²>1(n∈N*,n>

用数学归纳法证明1/n+1/(n+1)+1/(n+2)+……1/n²>1(n∈N*,n>1)
用数学归纳法证明1/n+1/(n+1)+1/(n+2)+……1/n²>1(n∈N*,n>1)

用数学归纳法证明1/n+1/(n+1)+1/(n+2)+……1/n²>1(n∈N*,n>1)
n=2略
n=k时有1/k+1/(k+1)+……+1/k²>1
k≥2
令a=1/k+1/(k+1)+……+1/k²>1
则n=k+1
1/(k+1)+1/(k+2)+……+1/(k+1)²
=a-1/k+1/(k²+1)+……+1/(k+1)²
因为1/(k²+1)>1/(k+1)²
1/(k²+2)>1/(k+1)²
……
所以a-1/k+1/(k²+1)+……+1/(k+1)²>a-1/k+1/(k+1)²+……+1/(k+1)²
=a-1/k+(2k+1)*1/(k+1)²
=a+(2k²+k-k²-2k-1)/k(k+1)²
=a+[(k-1/2)²-5/4]k(k+1)²
k≥2
所以a+[(k-1/2)²-5/4]k(k+1)²>a>1
所以n=k+1
1/(k+1)+1/(k+2)+……+1/(k+1)²>1
综上,……