Question

已知f(n)=1+1/2+1/3+...+1/n 用数学归纳法证明f(2^n)>n/2时,f(2^(k+1))-f(2^k)=?

Answer 1

n=1时，f(2)=1+1/2>1
假设当n=k时成立，下证当n=k+1时也成立
f(2^(k+1))=f(2^k)+1/(2^k+1)+1/(2^k+2)+...+1/(2^(k+1))
>k/2+1/(2^k+1)+1/(2^k+2)+...+1/(2^k+2^k))
注：(2^k+2^k)=2*2^k=2^(k+1)
从第二项起每项都用最后一项代替
>k/2+1/2^(k+1)+1/2^(k+1)+...+1/2^(k+1)
=k/2+2^k/2^(k+1)
=k/2+1/2
=(k+1)/2
不等式成立

Answer 2

f(2^(k+1))-f(2^k)
=
1/(2^k+1)+1/(2^k+2)+...+1/(2^(k+1))
按照题目就是这个结果

Answer 3

1.
当n=1时，f(2^1)=f(2)=1+1/2=3/2,n/2=1/2,f(2^n)>n/2成立；
2.
假设当n=k(k>=0且k为整数)时
f(2^n)>n/2成立，即
f(2^k)>k/2,则当
n=k+1时，有
f(2^(k+1))=f(2^k)+1/(2^k+1)+1/(2^k+2)+...+1/(2^(k+1))>k/2+1/(2^k+1)+1/(2^k+2)+...+1/(2^(k+1))；则要证f(2^(k+1))>(k+1)/2，只要证k/2+1/(2^k+1)+1/(2^k+2)+...+1/(2^(k+1))>(k+1)/2,则只要证1/(2^k+1)+1/(2^k+2)+...+1/(2^(k+1))>1/2
1/(2^k+1)+1/(2^k+2)+...+1/(2^(k+1))>2^k*1/(2^(k+1))=1/2
说明：2^k个分式都大于1/(2^(k+1))，则它们的和大于2^k乘以1/(2^(k+1))
f(2^(k+1))-f(2^k)=1/(2^k+1)+1/(2^k+2)+...+1/(2^(k+1))>2^k*1/(2^(k+1))=1/2
综合1，2得对任意正整数n,f(2^n)>n/2