证:a(n+1)*b(n+1)=1/2(an+bn)*2an*bn/(an+bn)=an*bn
∴an*bn=……=a0*b0>0 (常数) 令a0*b0=k
∴a(n+1)=1/2(an+bn)=≥k^½ (常数)
b(n+1)=2an*bn/(an+bn)=2k/(an+bn)≤k^½ (常数) 【an=bn时,取等号】
而a(n+1)-b(n+1)=1/2(an+bn)-2an*bn/(an+bn)=(an-bn)²/[2(an+bn)]≥0
即a(n+1)≥b(n+1)也就是an≥bn
∴a(n+1)=1/2(an+bn)≤1/2(an+an)=an
b(n+1)=2an*bn/(an+bn)=2/[1/an+1/bn)]≥2/[1/bn+1/bn)]=bn
即数列{an}单调递减,即数列{bn}单调递增。
∴数列{an},{bn}收敛
追问那an bn 的极限怎么求呢