求lim(1-1/4)(1-1/9).(1-1/n^2) n趋于无穷

来源：学生作业学帮网编辑：学帮网时间：2024/05/13 01:32:48

1-1/n^2=(n-1)/n*(n+1)/n
(1-1/4)(1-1/9).(1-1/n^2)
=1/2*3/2*2/3*4/3*3/4*5/4*……*(n-1)/n*(n+1)/n 中间项约去
＝(n+1)/2n
lim(1-1/4)(1-1/9).(1-1/n^2) n趋于无穷
=1/2

1-1/n^2=(n^2-1)/n^2=(n+1)(n-1)/n^2
所以
（1-1/4)(1-1/9)....(1-1/n^2)
=[1*3/2/2]*[2*4/3/3]*[3*5/4/4]*.....*[(n-1)(n+1)/n/n]
=1/2*(n+1)/n
->1/2（n->无穷)

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