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^

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