观察下列不等式:1>1/2,1+1/2+1/ 3>1,
2个回答

1+1/2+1/3+...+1/(2^n-1)>n/2

现在归纳证明

n=1时,1>1/2成立

现在证明n=k时成立(k>1)

由归纳知1+1/2+1/3+..+1/(2^(k-1)-1)>(k-1)/2

而n=k时

1+1/2+1/3+...+1/(2^k-1)

=(1+1/2+1/3+..+1/(2^(k-1)-1))

+ ( 1/(2^(k-1))+1/(2^(k-1)+1)+...1/(2^k-1) )

>(k-1)/2+ ( 1/(2^(k-1))+1/(2^(k-1)+1)+...1/(2^k-1) )

对 1/(2^(k-1))+1/(2^(k-1)+1)+...1/(2^k-1) 进行缩放,可知有2^(k-1)项,每项都大于1/2^k,所以

1/(2^(k-1))+1/(2^(k-1)+1)+...1/(2^k-1)>2^(k-1)/2^k=1/2

所以合起来1+1/2+1/3+...+1/(2^k-1)>(k-1)/2+1/2=k/2

得证