Problem 62 - Essential Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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 \(\infty\).

Step 4: 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.