利用:a1=1及a(n+1)=2an/(2+an),得:
a1=1
a2=2/3
a3=1/2=2/4
a4=2/5
猜测:an=2/(n+1)
证明:
1、当n=1时,an=a1=2/(1+1)=1,满足;
2、设:当n=k时,ak=2/(k+1)
则当n=k+1时,
a(k+1)=2ak/(2+ak) 【以ak=2/(k+1)代入】
=2/[(k+1)+1]
即当n=k+1时也成立
从而得证.
bn=an/n=2/[n(n+1)]=2[(1/n)-1/(n+1)]
则:
Sn=2{[(1/1)-(1/2)]+[(1/2)-(1/3)]+[(1/3)-(1/4)]+…+[1/n-1/(n+1)]}
=(2n)/(n+1)