Use the Direct Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftyfrac{sin^2n}{n^3}

CheemnCatelvew 2021-01-13 Answered
Use the Direct Comparison Test to determine the convergence or divergence of the series.
n=1sin2nn3
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Expert Answer

Obiajulu
Answered 2021-01-14 Author has 98 answers
Given:
The series, n=1sin2nn3
To determine the convergence or divergence of the series using the Direct Comparison Test.
The Direct Comparison Test:
Let, xn0,yn0 for all nN and xnynn. Then (i) if yn is convergent xn is convergent
(ii) If xn is divergent yn is divergent
Let, n=1sin2(n)n3
Here, consider xn=sin2n and yn=1n3
n=1sin2(n)n3=n=1|sin2(n)n3|
=n=1|sin2(n)|n3
=n=1|sin(n)|2n3
=n=11n3
That is, |sin(n)|2n31n3
And by p-series test, the series n=11n3 is convergent.
Therefore, by the Direct Comparison test xn is convergent.
Thus, by using the Direct Comparison test the series, n=1sin2(n)n3 is absolutely convergent
Hence, the given series is convergent.
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Jeffrey Jordon
Answered 2021-12-27 Author has 2262 answers

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