点（n,Sn/n）都在函数f(x)=x+an/2x
=》Sn/n=n+an/(2n)
=>Sn=n^2+an/2
an=Sn-S(n-1)=(n^2-(n-1)^2)+(an-a(n-1))/2
=>an=2n-1+an/2-a(n-1)/2
=>an+a(n-1)=4n-2 (a)
=>a(n-1)+a(n-2)=4n-6 (b)
=>an-a(n-2)=4 (i)
又a1=S1=1+a1/2
=>a1=2 (ii)
S2=a1+a2=4*2-2=6
=>a2=4 (iii)
(i）（ii)(iii)=>
=>an=2n
g(n)=(1+2/an)^n=(1+1/n)^n
最后就是归结到证明2