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

sagnuhh 2021-02-11 Answered
Explore the convergence of the series:
n=1narctan2(π8n+1)
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Expert Answer

Aniqa O'Neill
Answered 2021-02-12 Author has 100 answers

Given, the series n=1n[tan1(π8n+1)]2

We will use limit comparison test to test the convergence of it.

Limit Comparison Test:

For the two series n=1an and n=1bn with an0,bn>0,n.

Define c=limnanbn.

If 0, then both the series converge or diverge together.

In our case, an=n[tan1(π8n+1)]2

We know that, tan1(1x)1x for large values of x.

So, we have

tan1(π8n+1)1n

(tan1(π8n+1))21n2

n(tan1(π8n+1))2nn2=1n

This suggests we consider bn=1n whose series n=11n diverges.

By Limit comparison test,

limnanbn

=limnn[tan1(π8n+1)]21n

=limnn2[tan1(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=1bn diverges, the series

=n=1n[tan1(π8n+1)]2 must diverge.

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Jeffrey Jordon
Answered 2021-12-27 Author has 2064 answers

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