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

Question

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

brawnyN · Accepted Answer

Given:∑n=1∞1n+ln⁡nTo prove:The convergence or divergence of the given infinite series by using the limit comparison test.Limit comparison test:For series ∑n=1∞an and ∑n=1∞bn, where an,bn&amp;gt;0 for all n.If limn→∞(anbn)=c, 0 then either both series converges or both series diverges.Consider,∑n=1∞1n+ln⁡nHere, an=1n, bn=1n+ln⁡n⇒an=1n&amp;gt;0, bn=1n+ln⁡n&amp;gt;0Then,limn→∞(anbn)=limn→∞((1n)(1n+ln⁡n))=limn→∞(1n×(n+ln⁡n))=limn→∞(n+ln⁡nn)=limn→∞(nn+ln⁡nn)=limn→∞(1+ln⁡nn)=limn→∞1+limn→∞ln⁡nn=limn→∞(anbn)=1+limn→∞ln⁡nnHere to find the limn→∞ln⁡nnlimn→∞ln⁡nn is undefined form ∞∞Therefore, apply L&#039;Hospital&#039;s rule,limx→∞f(x)g(x)=limx→∞f′(x)g′(x)⇒limn→∞ln⁡nn=limx→∞(1n)(12n)=limn→∞(1n×2n)=limn→∞(2n)Applying the infinity property: limx→∞cxa=0limn→∞ln⁡nn=0Therefore, equation (1) becomes,limn→∞(anbn)=1+0=1⇒limn→∞(anbn)=1

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=1}^inftyfrac{1}{sqrt n+ln n}

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