高三数学数列问题~

来源：百度知道 编辑：UC知道 时间：2024/05/12 00:57:35
已知数列{an}{bn}分别满足a1a2a3…an=n!,b1+b2+b3...+bn=an^2
(1)求数列{an}{bn}的通项公式
（2）若数列{1/[(bn·b(n+1)]}的前n项和为Sn,求lim(x趋向无穷）[（3/2）Sn]的值

(1)an=a1a2a3…an/[a1a2a3…a(n-1)]=n!/(n-1)!=n
bn=(b1+b2+b3...+bn)-[b1+b2+b3...+b(n-1)]
=n^2-(n-1)^2
=2n-1

(2)(bn·b(n+1)=(2n-1)[2(n+1)-1]=(2n-1)(2n+1)
1/[(bn·b(n+1)]=1/[(2n-1)(2n+1)]
=(1/2)*{[1/(2n-1)-[1/(2n+1)]}

Sn=(1/2)*{(1-1/3)+(1/3-1/5)+……+[1/(2n-1)]-[1/(2n+1)]}
=(1/2)*[1-1/(2n+1)]

lim(x趋向无穷）[（3/2）Sn]
=(3/2)*(1/2)lim(x趋向无穷）[1-1/(2n+1)]
=3/4