Explore the convergence of the series: ∑n=1∞n⋅arctan2⁡(π8n+1)

Question

Explore the convergence of the series:  ∑n=1∞n⋅arctan2⁡(π8n+1)

Aniqa O&#039;Neill · Accepted Answer

Given, the series ∑n=1∞n⋅[tan−1⁡(π8n+1)]2  We will use limit comparison test to test the convergence of it.  Limit Comparison Test:  For the two series ∑n=1∞an and ∑n=1∞bn with an≥0,bn&amp;gt;0,∀n.  Define c=limn→∞anbn.  If 0, then both the series converge or diverge together.  In our case, an=n[tan−1⁡(π8n+1)]2  We know that, tan−1⁡(1x)∼1x for large values of x.  So, we have  tan−1⁡(π8n+1)∼1n  ⇒(tan−1⁡(π8n+1))2∼1n2  ⇒n(tan−1⁡(π8n+1))2∼nn2=1n  This suggests we consider bn=1n whose series ∑n=1∞1n diverges.  By Limit comparison test,  limn→∞anbn  =limn→∞n[tan−1⁡(π8n+1)]21n  =limn→∞n2[tan−1⁡(π8n+1)]2  We used WolframAlpha calculator to obtain its value,  =π264  Hence we have c=π264 which lies between 0 and ∞  Therefore, by the limit comparison test, as ∑n=1∞bn diverges, the series  =∑n=1∞n⋅[tan−1⁡(π8n+1)]2 must diverge.

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

Explore the convergence of the series: sum_{n=1}^infty ncdotarctan^2left(frac{pi}{8n+1}right)

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