An/森改Bn=[(A1+A2n-1)/2]/[(B1+B2n-1)/2]=[(A1+A2n-1)n/2]/[(B1+B2n-1)n/2]=S2n-1/T2n-1=2*(2n-1)/[3(2n-1)+1]=(4n-2)/(6n-2)当n→+∞时An/此锋判Bn=4/6=2/3即为所求基枝
2/3
Sn/Tn=2n/(3n+1)可令Sn=2n^2Tn=n(3n+1)用Sn和Tn表示出An和码让槐Bn再求极限滑拿,不是迟友2/3