记S(2^k)=1+1/2+1/3+1/4+...+1/2^k
当 2^(t-1)<m<=2^t时, 1/m>=1/2^t
这步懂吗?

@三江方士
对每一项放缩,所以
S(2^k)=
1+1/2+1/3+1/4+1/5+/1/6+1/7+1/8+1/9+1/10+1/11+1/12+1/13+1/14+1/15+...+1/2^k
>=
1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+1/16+1/16+1/16+1/16+1/16+1/16+1/16+...+1/2^k
=1+(1/2)*k