Problem 33 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} n \sin \frac{1}{n} $$

Step-by-Step Solution

Answer

The series diverges by the divergence test since the limit of the general term is not zero.

Step 1: Identify the Behavior of the General Term

We start by analyzing the general term of the given series \( a_n = n \sin \frac{1}{n} \). By using the approximation \( \sin x \approx x \) for small \( x \), we have \( \sin \frac{1}{n} \approx \frac{1}{n} \). So, \( a_n \approx n \cdot \frac{1}{n} = 1 \).

Step 2: Analyze the Limit of the General Term

Find \( \lim_{n \to \infty} a_n \). Since \( a_n = n \sin \frac{1}{n} \approx 1 \), the limit is \( \lim_{n \to \infty} a_n = 1 \).

Step 3: Apply the Divergence Test

The divergence test states that if \( \lim_{n \to \infty} a_n eq 0 \), then the series \( \sum a_n \) diverges. From Step 2, we have \( \lim_{n \to \infty} a_n = 1 eq 0 \).

Step 4: Conclusion

Since \( \lim_{n \to \infty} a_n = 1 eq 0 \), apply the divergence test to conclude that the series \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \) diverges.

Key Concepts

General Term Analysis

When studying a series, the very first step is always to look at the general term of the series. Understanding how each term behaves individually can give us significant hints about the convergence or divergence of the entire series.
For the given series \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \), we can define the general term as \( a_n = n \sin \frac{1}{n} \). In mathematics, small inputs to the sine function approximate the input itself, i.e., \( \sin(x) \approx x \) for small \( x \).
With \( \sin \frac{1}{n} \approx \frac{1}{n} \) for sufficiently large \( n \), this simplifies our general term analysis:

• \( a_n = n \cdot \frac{1}{n} = 1 \)

This simplification provides us with an essential clue for the next step in our analysis. As \( n \) becomes very large, each term \( a_n \) closely resembles the constant value 1.

Divergence Test

Understanding how the divergence test works is crucial when analyzing series. The divergence test, also known as the nth-term test, provides a straightforward way to determine if a series diverges.
According to this test, if the limit of the general term \( \lim_{n \to \infty} a_n eq 0 \), the series must diverge. This implies that when the terms of a series do not approach zero, the series cannot sum to a finite value.
In our case:

• The general term is \( a_n = n \sin \frac{1}{n} \)
• We found that \( \lim_{n \to \infty} a_n = 1 \)

Because 1 is not equal to zero, the divergence test tells us that the series \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \) diverges. This test is a powerful tool because it can quickly identify divergent series when a limit clearly doesn’t approach zero.

Limit of a Sequence

The concept of limits plays a vital role in understanding the behavior of sequences and series. The limit of a sequence describes what the terms approach as the sequence progresses towards infinity.
For the sequence of terms \( a_n = n \sin \frac{1}{n} \), we calculated the limit as
\[ \lim_{n \to \infty} a_n = 1 \]
This tells us that each term in our sequence approaches 1 as \( n \) becomes very large.
The importance of finding this limit lies in its use with convergence tests. If the limit of a sequence's terms is non-zero, as it is in this case, the series diverges as per the divergence test. Therefore, understanding limits is crucial for determining the overall behavior of series and their convergence properties.