用数学归纳法证明1∕n+1∕（n+1)+1∕(n+2)+…+1∕（n的平方）〉1,其中n〉1

Question

用数学归纳法证明1∕n+1∕（n+1)+1∕(n+2)+…+1∕（n的平方）〉1,其中n〉1

Answer 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,……