Prove that the sequence sin &#x2061;<!-- ⁡ --> ( k )

Jorge Lawson

Jorge Lawson

Answered question

2022-05-18

Prove that the sequence sin ( k ) + k 2 1 + 3 k 4 converges using the ϵ criteria

Answer & Explanation

asafand2c

asafand2c

Beginner2022-05-19Added 11 answers

First, noting that | sin ( k ) | 1, then given ϵ > 0 we have
k > K 1 = 2 ϵ
whenever, k>K1=2ϵ−−√.
Next, we have
| k 2 1 + 3 k 4 1 3 | = | 1 ( 3 k 2 + 1 + 3 k 4 ) 3 1 + 3 k 4 | 1 k 4 < ϵ 2
whenever k > K 2 = 2 ϵ 4
Finally, using the triangle inequality, we see that given ϵ > 0,
| sin ( k ) + k 2 1 + 3 k 4 1 3 | | sin ( k ) 1 + 3 k 4 | + | k 2 1 + 3 k 4 1 3 | < ϵ 2 + ϵ 2 = ϵ
whenever k > max ( K 1 , K 2 )

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