Determine whether the series ∑an an converges or diverges: Use the Alternating Series Test. ∑n=2∞(−1)nnln⁡(n)

Question

Determine whether the series ∑an an converges or diverges: Use the Alternating Series Test.  ∑n=2∞(−1)nnln⁡(n)

Obiajulu · Accepted Answer

Consider the series ∑n=2∞(−1)nnln⁡(n)  Take an=(−1)nnln⁡(n) and bn=nln⁡(n)  The Alternating series test is stated below:  Suppose the series ∑an and either an=(−1)nbn or an=(−1)n+1bn where bn≥0 for all n. Then if,  1.limn→∞bn=0  2. {bn} is a decreasing sequence  The series  is convergent.  Check the first condition for series convergent.  limn→∞bn=limn→∞(nln⁡(n))  =∞∞  The value of the limit is in the indeterminate form.  Apply LHopitals rule to find the limit as follows.  limn→∞bn=lim(ddn(n)ddn(ln⁡(n)))  =limn→∞(11n)  =limn→∞(n)  =∞  Observe that, the limit of the sequence goes to infinity as x goes to infinity. Thus, the series does not converges.  The divergence test states that, &quot;If limn→∞an≠0 then the series ∑an will diverge&quot;.  Since limn→∞bn≠0, the limit of the function an=(−1)nnln⁡(n) also not equal to zero as x goes to zero. That is, limn→∞an≠0.  By divergence test, it is concluded that the  alternating series ∑n=2∞(−1)nnln⁡(n) diverges.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Determine whether the series sum a_n an converges or diverges: Use the Alternating Series Test. sum_{n=2}^infty(-1)^nfrac{n}{ln(n)}

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