Asked by kelly
determine whethere the sequence a(n)=nsinn/n^3 converges or diverges. Explain why it does or does not converge. If it converges find the limit.
Sin(n) is always between -1 and 1. So, you have:
-1/n^2 < a(n) < 1/n^2
Both the lower bound and the upper bound converge to zero, so the sequence converges to zero.
Sin(n) is always between -1 and 1. So, you have:
-1/n^2 < a(n) < 1/n^2
Both the lower bound and the upper bound converge to zero, so the sequence converges to zero.
Answers
There are no human answers yet.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.