Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series. ∑n=2∞1nln⁡n

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ensojadasH · Accepted Answer

Apply integral test and check weather series is convergeny or divergent.  If the term series an can be represented by a positive, decreasing, continuous functionm f(n) then,  If limR→∞∫aRf(x)dx exists then ∑n=1∞an converges.  If limR→∞∫aRf(x)dx does not exist, then ∑n=1∞an diverges.  Consider the given series:  Sn=∑n=2∞1nln⁡n  Determine the convergence of the following Integral:  ∫2∞1xln⁡xdx=limt→∞∫2t1xln⁡xdx  Apply u-substitution: u=ln⁡(x)  ∫2t1xln⁡xdx=∫ln⁡(2)ln⁡(t)1udu  =∫ln⁡(2)ln⁡(t)u−12du  =[u−12+1−12+1]ln⁡(2)ln⁡(t)  =[2u]ln⁡(2)ln⁡(t)  =2ln⁡(t)−2ln⁡(2)  Find whether series diverges.  Applying Limits:  limt→∞∫2t1xln⁡xdx=limt→∞[2ln⁡(t)−2ln⁡(2)]  =[∞−2ln⁡(2)]  =∞  =Diverges  Hence,

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.sum_{n=2}^inftyfrac{1}{nsqrt{ln n}}

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