请教数学达人解决一个数学难题

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

1+(1/2)+(1/3)+.........(1/2^n) >(n/2)+1
(lnx)'=1/x
1+(1/2)+(1/3)+.........(1/2^n)
=1+[(1/2)+(1/3)+.........(1/2^n)]
>(从1到2^n)∫(1/x)
=ln(2^n)-ln1
=nln2
>1+n/2
n>=6时，一定成立
n=1,2,3,4,5容易验证成立。

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
1+(1/2)+(1/3)+.........(1/2^n) >(n/2)+1
(lnx)'=1/x
1+(1/2)+(1/3)+.........(1/2^n)
=1