Problem 23 - Calculus: Early Transcendentals

Question

Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} \tan(1/n) \)

Step-by-Step Solution

Answer

The series \( \sum_{n = 1}^{\infty} \tan(1/n) \) diverges.

Step 1: Initial Assessment

We have the series \( \sum_{n=1}^{\infty} \tan(1/n) \). Our first step is to consider the behavior of \( \tan(1/n) \) as \( n \to \infty \). Since \( 1/n \to 0 \) as \( n \to \infty \), \( \tan(1/n) \approx 1/n \). This suggests that each term behaves like \( 1/n \).

Step 2: Comparison to Known Series

We compare \( \tan(1/n) \) to the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \). The harmonic series is known to diverge. For large \( n \), \( \tan(1/n) \approx 1/n \), so the terms of the series \( \tan(1/n) \) behave similarly to those of the harmonic series.

Step 3: Convergence Test

Since the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges, and \( \tan(1/n) \approx 1/n \) for large \( n \), we use the Comparison Test. Since \( \tan(1/n) \geq \frac{1}{n} \) for sufficiently large \( n \), the divergence of the harmonic series implies the divergence of \( \sum_{n=1}^{\infty} \tan(1/n) \).

Step 4: Conclusion

By the Comparison Test, because \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges and \( \tan(1/n) \geq \frac{1}{n} \) for large \( n \), the series \( \sum_{n=1}^{\infty} \tan(1/n) \) also diverges.

Key Concepts

Harmonic Series

The harmonic series is a classic example used in studying the convergence or divergence of series. It is defined as:

• \( \sum_{n=1}^{\infty} \frac{1}{n} \)

This series comes under a fundamental category of infinite series that diverges. In simpler terms, even as you keep adding smaller and smaller fractions as seen with \( \frac{1}{n} \), the entire sum keeps growing without bound.
The harmonic series diverges because it grows slower than a geometric series but is significant enough to never level off completely.
For students, it's important to recognize that any series which behaves similarly to the harmonic series might also diverge. This understanding is crucial when applying tests like the Comparison Test.

Comparison Test

The Comparison Test is a valuable tool for determining if a series converges or diverges by comparing it to another series whose behavior is known.

• Consider two series \( \sum a_n \) and \( \sum b_n \)
• If \( 0 \leq a_n \leq b_n \) for all \( n \) beyond some number, and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.
• Conversely, if \( \sum b_n \) converges, so too must \( \sum a_n \).

In our exercise, we used the Comparison Test to compare \( \tan(1/n) \) to \( 1/n \) because for sufficiently large \( n \), \( \tan(1/n) \approx 1/n \). With the harmonic series still diverging, and \( \tan(1/n) \) behaving similarly or being greater, it too must diverge. This illustrates how the behavior of one series can predict the behavior of another similar series.

Divergence

Divergence refers to a series where the sum does not settle at a certain number but keeps increasing indefinitely.

• If the sum of series \( \sum a_n \) grows without bound, it is said to diverge.
• This usually means, as adding terms progresses, it fails to approach any finite limit.

Understanding divergence helps in identifying when an infinite process doesn't stabilize or reach a specific value. In the given exercise, the divergence of the series \( \sum_{n=1}^{\infty} \tan(1/n) \) was established through its comparison to the well-known divergent harmonic series.
Even though the terms \( \tan(1/n) \) are connected to a trigonometric function, their behavior in the limits makes them behave like terms of a divergent series.