Convergence of sequence

👨🏻‍🏫 學術台數學

[匿名]

2013-09-27 3:34 am

Show the sequence a_n = (1/n)sin 2n converges and hence find the limit.

更新1:

n->infinity for the limit

回答 (1)

myisland8132

2013-09-29 7:39 pm

✔ 最佳答案

a_n = sin(2n)/n

Let L = 0, N = 1/ϵ, when n > N

|a_n - L|

= |sin(2n)/n - 0|

<= 1/n

< 1/N

< ϵ

So, sequence a_n = (1/n)sin 2n converges and the limit is 0

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