when n=2,we have,
f(4)=1+1/2+1/3+1/4=25/12>(2+2)/2=2
Assume when n=k,we have f(2^k)>(k+2)/2,then,
when n=k+1,f[2^(k+1)]=f(2^k)+1/(2^k+1)+1/(2^k+2)+...+1/(2^k +2^k) > (k+2)/2 + 2^k * 1/(2^k +2^k) = [(k+1)+2]/2
So when n>1,we have f(2^n)>(n+2)/2