Use the Limit Comparison Test to determine the convergence or divergence of the series. ∑n=1∞2n2−13n5+2n+1

Question

Use the Limit Comparison Test to determine the convergence or divergence of the series.  ∑n=1∞2n2−13n5+2n+1

irwchh · Accepted Answer

We have to check the given series ∑n=1∞2n2−13n5+2n+1 is convergent or divergent using the limit comparision test. According to limit comparision test if two series ∑n=1∞an and ∑n=1∞bn with an&amp;gt;0,bn&amp;gt;0 for all n. Then if limn→∞anbn=c with 0 then either both series converges or both series divergent. Let ∑n=1∞an=∑n=1∞an2n2−13n5+2n+1 Take common highest power n from denominator and numerator we get ∑n=1∞2n2−13n5+2n+1=∑n=1∞n2(2−1n2)n5(3+2n4+1n5) =∑n=1∞(2−1n2)n3(3+2n4+1n5) So, ∑n=1∞an=∑n=1∞(2−1n2)n3(3+2n4+1n5) Let another series bn=1n3 bn is convergent p-series since p=3 Now, limn→∞anbn=limn→∞(2−1n2)n3(3+2n4+1n5)1n3 =limn→∞(2−1n2)(3+2n4+1n5) =2−03+0+0 =23 =limn→∞anbn=23 which is finite and positive. Therefore we can conclude by limit comparison test series ∑n=1∞an will be convergent. So 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{2n^2-1}{3n^5+2n+1}

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