Determine whether the series converges or diverges. ∑n=2∞1nln⁡n

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Determine whether the series converges or diverges.  ∑n=2∞1nln⁡n

Ayesha Gomez · Accepted Answer

Given that: The series ∑n=2∞1nln⁡n By using, Limit - Comparison test : Suppose that we have two series ∑an and ∑bn with an≥0,bn&amp;gt;0 for all n Define c=limn→∞anbn If c is positive (that is c&amp;gt;0) and is finite then either both series converges or both series diverge. P - series test: The series is of the form ∑1np is converges if p&amp;gt;1 and diverges if 0&amp;lt;p≤1 Let , The original series is an=1nln⁡n We need to find the series similar to the original series but simpler. 1nln⁡n&amp;lt;1n Then, The comparison series is bn=1n By using the p- series test, ∑n=2∞1n=∑n=2∞1n1 Here p = 1 Then, By the p - series test bn is divergent. c=limn→∞anbn =limn→∞1nln⁡n×n1 =limn→∞1ln⁡n =diverge. Then, By the Limit - Comparison Test, The series ∑n=2∞1nln⁡n is diverges. Therefore, The series ∑n=2∞1nln⁡n is diverges.

Determine whether the series converges or diverges. sum_{n=2}^inftyfrac{1}{nln n}

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