Use the Limit Comparison Test to determine the convergence or divergence of the series. ∑n=1∞sin⁡1n

Question

Use the Limit Comparison Test to determine the convergence or divergence of the series.  ∑n=1∞sin⁡1n

2abehn · Accepted Answer

Limit comparison test:  Suppose we have two series ∑an and ∑bn. Define  limn→∞anbn=L  If L is positive and finite then either both series convergent or both series divergent.  Given that,  ∑n=1∞sin⁡1n  Since we have,  ∑n=1∞sin⁡1n  =∑n=1∞(1n−(1n)313!+(1n)515!+...)  [Using sin⁡x=x−x33!+x55!+...]  Here  an=(1n−(1n)313!+(1n)515!+...)  So  bn=1n  Now evaluate limn→∞anbn, so  limn→∞anbn=limn→∞1n−(1n)313!+(1n)515!+...1n  =limn→∞1n[1−(1n)213!+(1n)415!+...]1n  =limn→∞[1−(1n)213!+(1n)415!+...]  =1  So limn→∞anbn=1&amp;gt;0 which is positive and finite.  Using p series test [∑1np is convergent if p&amp;gt;1 or divergent if p≤1], ∑bn is divergent.  Since limn→∞anbn=1&amp;gt;0 is positive and finite. ∑bn is divergent, so using limit comparison test ∑an is also divergent.  Hence ∑n=1∞sin⁡1n is divergent.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Use the Limit Comparison Test to determine the convergence or divergence of the series. sum_{n=1}^inftysinfrac{1}{n}

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