Problem 45 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \sin \frac{1}{n} $$

Step-by-Step Solution

Answer

The series diverges, similar to the harmonic series, by the limit comparison test.

Step 1: Analyze the General Term

Consider the general term of the series, which is \( \sin \frac{1}{n} \). For very large \( n \), \( \sin \frac{1}{n} \) can be approximated by its Taylor expansion for small angles: \( \sin x \approx x \). Therefore, \( \sin \frac{1}{n} \approx \frac{1}{n} \) when \( n \) is large.

Step 2: Comparison to the Harmonic Series

Recall that the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) is known to diverge. Since \( \sin \frac{1}{n} \approx \frac{1}{n} \) for large \( n \), the terms of our series are roughly comparable to the terms of the harmonic series.

Step 3: Use the Limit Comparison Test

Apply the limit comparison test between \( \sum_{n=1}^{\infty} \sin \frac{1}{n} \) and \( \sum_{n=1}^{\infty} \frac{1}{n} \). Compute the limit: \[ L = \lim_{{n \to \infty}} \frac{\sin \frac{1}{n}}{\frac{1}{n}} = \lim_{{n \to \infty}} \frac{\sin \frac{1}{n}}{\frac{1}{n}} = \lim_{{n \to \infty}} \frac{n \cdot \sin \frac{1}{n}}{1} \]Using L'Hôpital's rule or the small angle approximation, it simplifies to 1. Since \( L = 1 \) and the harmonic series diverges, the given series \( \sum_{n=1}^{\infty} \sin \frac{1}{n} \) also diverges by the limit comparison test.

Key Concepts

Harmonic Series

The harmonic series is a famous infinite series of the form: \[ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots \]Though it might seem that this series would converge because its terms become smaller, it actually diverges. The terms decrease in size, but not quickly enough. There isn't a point where the sum levels out. Every series that resembles the harmonic series needs careful analysis because of this unique behavior.

• Key aspect: The divergence occurs because the partial sums continue to grow without bound.
• Even if we slow the addition by further fractions, like all fractions being only slightly smaller than \( \frac{1}{n} \), the series still diverges.

Limit Comparison Test

The limit comparison test is a helpful method when determining the convergence or divergence of a series. It's particularly effective when you suspect one series behaves like another known series. Here's how it works:

• Take two series, \( \sum a_n \) and \( \sum b_n \).
• Compute the limit \( L = \lim_{n \to \infty} \frac{a_n}{b_n} \).
• If \( 0 < L < \infty \), then either both series converge, or both series diverge.

This method helps especially when the terms of the series we're examining are complex or unfamiliar. Comparing them to more manageable, well-known series like the geometric or harmonic series can provide a clear answer. When we analyzed \( \sum_{n=1}^{\infty} \sin \frac{1}{n} \), we used this test to compare it with the harmonic series. Because the limit was 1, indicating similar behavior, and because the harmonic series diverges, the original series also diverges.

Taylor Expansion

The Taylor expansion is a powerful tool for approximating functions. It's especially useful in series and calculus for simplifying complex expressions into more manageable forms. At the core, the Taylor expansion of a function about the point \( x = 0 \) (called a Maclaurin series) is:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \]This transforms complex functions into polynomial parts, which are easier to deal with.For small angles, such as in \( \sin x \), we use the expansion:\[ \sin x \approx x - \frac{x^3}{6} + \cdots \]For \( x = \frac{1}{n} \), the term \( \frac{x^3}{6} \) becomes negligible for large \( n \), so \( \sin \frac{1}{n} \approx \frac{1}{n} \). This approximation allows us to analyze the series \( \sum_{n=1}^{\infty} \sin \frac{1}{n} \) by comparing it to the harmonic series. Such a handy tool makes it simpler to understand why certain series behave like others.

Divergence of Series

Divergence refers to the behavior of a series that does not settle on a finite number, but instead tends to infinity as more terms are added. Understanding why a series diverges is crucial for determining its overall behavior.In many series, like the harmonic series, divergence occurs because the terms do not decrease fast enough. That means, despite each added term being smaller, the sum continues to grow.

• A diverging series can never reach a specific limit or cap.
• If you plot the partial sums of a diverging series, you'll see a trend upwards, without leveling out.

In the original exercise, using the limit comparison test clarified the divergence of the series \( \sum_{n=1}^{\infty} \sin \frac{1}{n} \). Since the terms \( \sin \frac{1}{n} \) mirror those of the divergent harmonic series, they similarly cause divergence. Recognizing divergence helps in predicting the behavior and outcome of a sequence or series.