a（n+1）=2an/（an+1）
1/a（n+1）=1/2(1/an+1)
1/a（n+1）-1=1/2(1/an-1)
[1/a（n+1）-1]/(1/an-1)=1/2 (1/a1-1)=3/2-1=1/2
1/an-1=1/2*(1/2)^(n-1)=1/2^n an=1/[1/2^n+1]=2^n/[2^n+1]
n/an=n/2^n+n（分项求和）
s1=1*1/2+2*1/2^2+……+n/2^n
2sn1=1*1/2^0+2*1/2^1+……+n/2^(n+1)
2sn1-sn1=sn1=1+1/2+1/4+……+1/2^(n+1)=1+1-1/2^(n+1)=2-1/2^(n+1)
sn2=1+2+3+……+n=(1+n)*n/2
sn=sn1+sn2=2-1/2^(n+1)+(1+n)*n/2=ok