Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the convergence or divergence of the series using other methods. ∑n=1∞n2(n+1)(n2+2)

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Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the convergence or divergence of the series using other methods.  ∑n=1∞n2(n+1)(n2+2)

StrycharzT · Accepted Answer

Ratio test: Let the series is ∑an then, limn→∞an+1an=L If L&amp;lt;1, series is convergent. If L&amp;gt;1, series is divergent. If L=1, may be convergent or divergent. Given that, ∑n=1∞n2(n+1)(n2+2) Here an=n2(n+1)(n2+2) so an+1=(n+1)2(n+2)((n+1)2+2) Now apply the ratio test, limn→∞an+1an=limn→∞((n+1)2(n+2)((n+1)2+2)n2(n+1)(n2+2)) limn→∞((n+1)2(n+2)((n+1)2+2)×(n+1)(n2+2)n2) limn→∞(n2(1+1n)2n3(1+2n)((1+1n)2+2n2)×n3(1+1n)(1+2n2)n2) limn→∞((1+1n)2(1+2n)((1+1n)2+2n2)×(1+1n)(1+2n2)1) =1 Since limn→∞an+1an=L, where L=1. So ratio test fails. Comparison test: Let the series is ∑an then, limn→∞anbn=L Where L is positive and finite then ∑an and ∑bn are both convergent or divergent. Given that, ∑n=1∞n2(n+1)(n2+2) Here an=n2(n+1)(n2+2) so bn=n2n3=1n Now using the comparison test, limn→∞anbn=limn→∞(n2(n+1)(n2+2)1n) =limn→∞(n2(n+1)(n2+2)×n) =limn→∞(n2n3(1+1n)(1+2n2)×n)

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the convergence or divergence of the series using other methods. sum_{n=1}^inftyfrac{n^2}{(n+1)(n^2+2)}

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