Use the Limit Comparison Test to prove convergence or divergence of the infinite series. ∑n=2∞nn3+1

Question

Use the Limit Comparison Test to prove convergence or divergence of the infinite series.  ∑n=2∞nn3+1

Arnold Odonnell · Accepted Answer

Let the given expression be represented as,  an=nn3+1  Consider another function bn=1n  According to the limit comparison test-  1. If limn→∞(anbn)= finite positive value, then series ∑n→∞an and ∑n→∞bn behave alike, i.e. either both converge or both diverge.  2. If limn→∞(anbn)=0 and ∑n→∞bn converges, then series ∑n→∞an will also converge.  3. If limn→∞(anbn)→∞ diverges, then series ∑n→∞an will also diverge.  Clearly the series ∑n=1∞bn is a converging series as the term limn→∞1n≈0  Check fot he series (anbn) as,  limn→∞(anbn)=(nn3−11n)  =limn→∞(nn3−1×n)  =limn→∞(n3n3−1)  =limn→∞(n3n3−1)  =limn→∞(11−1n3)  As n→∞, then 1n3→0,  limn→∞(anbn)=11−0  =1  Since limn→∞(anbn) is a finite positive value and ∑n=1∞bn is converging, therefore the limit comparison test implies that, the series ∑n=1∞an is also converging.  Consider the following,  ∑n=2∞nn3+1=∑n=1∞nn3+1−[nn3+1]n=1  =∑n=1∞nn3+1−113+1  =∑n=1∞nn3+1−12  Consider that the difference between two finite values is also finite.  Thus, it is concluded that the series =∑n=2∞nn3+1 is converging.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Use the Limit Comparison Test to prove convergence or divergence of the infinite series. sum_{n=2}^inftyfrac{n}{sqrt{n^3+1}}

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