Use the Direct Comparison Test to determine the convergence or divergence of the series. ∑n=1∞sin2⁡nn3

Question

Use the Direct Comparison Test to determine the convergence or divergence of the series.  ∑n=1∞sin2⁡nn3

Obiajulu · Accepted Answer

Given:  The series, ∑n=1∞sin2⁡nn3  To determine the convergence or divergence of the series using the Direct Comparison Test.  The Direct Comparison Test:  Let, xn≥0,yn≥0 for all n∈N and xn≤yn∀n. Then (i) if ∑yn is convergent ⇒∑xn is convergent  (ii) If ∑xn is divergent ⇒∑yn is divergent  Let, ∑n=1∞sin2⁡(n)n3  Here, consider xn=sin2⁡n and yn=1n3  ⇒∑n=1∞sin2⁡(n)n3=∑n=1∞|sin2⁡(n)n3|  =∑n=1∞|sin2⁡(n)|n3  =∑n=1∞|sin⁡(n)|2n3  =∑n=1∞1n3  That is, |sin⁡(n)|2n3≤1n3  And by p-series test, the series ∑n=1∞1n3 is convergent.  Therefore, by the Direct Comparison test ∑xn is convergent.  Thus, by using the Direct Comparison test the series, ∑n=1∞sin2⁡(n)n3 is absolutely convergent  Hence, the given series is convergent.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

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

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