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问题: 高中极限

设等差数列(an),(bn)的前n项和分别是Sn,Tn且满足Sn/Tn=(3n+1)/(5n+2),则当n趋向于无穷大时,an/bn无限趋向于?
A,4/7 B,3/5 C,1/2 D,1

解答:

设等差数列{an},{bn}的前n项和分别是Sn,Tn,且满足Sn/Tn=(3n+1)/(5n+2),则当n→∞时,an/bn无限趋向于?

S(2n-1) = [a1+a(2n-1)]•(2n-1)/2 = an•(2n-1)
T(2n-1) = [b1+b(2n-1)]•(2n-1)/2 = bn•(2n-1)
--->an/bn = S(2n-1)/T(2n-1)
     = [3(2n-1)+1]/[5(2n-1)+2] = (6n-2)/(10n-3)
n→∞时,an/bn = (6-2/n)/(10-3/n) = 3/5