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Chapter 7: Problem 62

Use the Limit Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \tan \frac{1}{n} $$

Short Answer

Expert verified
The given series \(\sum \tan(1/n)\) is divergent.

Step by step solution

01

Know the Limit Comparison Test (LCT)

Before starting with the problem, one must understand the Limit Comparison Test, which states that if we've two series \(\sum a_n\) and \(\sum b_n\) (where \(b_n > 0\) for all \(n\)), and the limit as \(n\) approaches infinity of \(a_n/b_n = L\), where \(0 < L < \infty\), both series either converge or diverge.
02

Choose a series to compare

For the given series, we observe that the term \(\tan (1/n)\) behaves similar to \(1/n\) as \(n\) approaches infinity as tanx is similar to \(x\) for small \(x\). Therefore, we'll compare the given series with the series \(\sum (1/n)\). We know that the series \(\sum (1/n)\) is a harmonic series which is divergent.
03

Apply LCT and compute limit

Apply the Limit Comparison Test (LCT) on both series. We evaluate:\[\lim_{{n \to \infty}} \frac{{\tan(1/n)}}{{1/n}}\]By using L'Hopital's rule (since the above form is 0/0), we get:\[\lim_{{n \to \infty}} \frac{{\sec^2(1/n) * (-1/n^2)}}{{-1/n^2}} = \lim_{{n \to \infty}} \sec^2(1/n)\]The limit as n approaches infinity \(1/n\) approaches 0, and \(\sec^2(0) = 1\), which is not equal to 0 or \(\infty\).
04

Convergence or Divergence

By the Limit Comparison Test, since the limit is finite and positive, the given series behaves the same way as the selected series, i.e., \(\sum (1/n)\). Since the selected series \(\sum (1/n)\) is divergent (harmonic series), the given series \(\sum \tan(1/n)\) is also divergent.

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