Use the Limit Comparison Test to determine the convergence or divergence of the series. ∑n=1∞1n2(n2+4)

Question

Use the Limit Comparison Test to determine the convergence or divergence of the series.  ∑n=1∞1n2(n2+4)

okomgcae · Accepted Answer

We have to find the given series is convergence or divergence. Limit comparison test: Suppose that we have two series ∑n=1∞an and ∑n=1∞bn with an≥0 and bn&amp;gt;0 for all n Then, if limn→∞anbn=c, with 0 then either both series convergent or both series divergent. We have given ∑n=1∞1n2(n2+4) Let ∑n=1∞an=∑n=1∞1n2(n2+4) Then, an=1n2(n2+4) And suppose ∑n=1∞bn=∑n=1∞1n4 another series. Then, bn=1n4 And ∑n=1∞bn is a p-series and p=4 so, it is convergent. Now, limn→∞anbn=limn→∞1a2(a2+4)1a4 =limn→∞1n2(n2+4)×n41 =limn→∞n4n4(1+4a2) =limn→∞1(1+4a2) =1 (finite and non zero) Since, series ∑n=1∞bn is convergent so, by comparison test series ∑n=1∞an is convergent. Therefore, given series is convergent.

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}^inftyfrac{1}{n^2(n^2+4)}

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