Problem 35
Question
Which of the series in Exercises 13 46 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
Verified Answer
The series diverges by the Comparison Test.
1Step 1: Rewrite the Series
Let's consider the series \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \). To better understand this series, we rewrite \( \sin \frac{1}{n} \) using the approximation \( \sin x \approx x \) for small values of \( x \). This gives us \( n \sin \frac{1}{n} \approx n \cdot \frac{1}{n} = 1 \). Thus, the terms of the series can be approximately simplified to 1, making the series \( \sum_{n=1}^{\infty} 1 \).
2Step 2: Analyze the Simplified Series
Now, the series \( \sum_{n=1}^{\infty} 1 \) is a very well-known divergent series because it represents the sum of the constant 1, repeated for every term in the series from 1 to infinity. The sum of an infinite number of ones is infinity, indicating that the series diverges.
3Step 3: Justification by Comparison Test
To firmly establish the divergence, we apply the Comparison Test. The terms \( a_n = n \sin \frac{1}{n} \) satisfy \( \sin \frac{1}{n} < \frac{1}{n} \). Hence, \( n \sin \frac{1}{n} > n \cdot \frac{1}{n} = 1 \). Since \( \sum_{n=1}^{\infty} 1 \) diverges, by the Comparison Test, \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \) also diverges.
Key Concepts
Comparison TestDivergent SeriesPower Series
Comparison Test
The Comparison Test is a method used to determine the convergence or divergence of series. It involves comparing a series with known behavior to another series with similar terms.
The general idea is if you have a series \( \sum a_n \), you compare it to another series \( \sum b_n \) where each term \( a_n \) is less than or equal to \( b_n \), or vice versa. By doing so, you can infer the behavior of the series based on the more intuitive one.
For example, if \( \sum b_n \) is a known divergent series and \( a_n \geq b_n \) for all \( n \), then \( \sum a_n \) also diverges. Conversely, if \( \sum b_n \) converges and \( a_n \leq b_n \), then \( \sum a_n \) converges as well.
This test is beneficial when deploying it alongside approximations, as seen in the problem where we simplify the behavior of \( n \sin \frac{1}{n} \). By knowing \( \sin \frac{1}{n} < \frac{1}{n} \), we conclude \( n \sin \frac{1}{n} > 1 \). Therefore, seeing \( \sum_{n=1}^{\infty} 1 \) diverges, \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \) also diverges according to the Comparison Test.
The general idea is if you have a series \( \sum a_n \), you compare it to another series \( \sum b_n \) where each term \( a_n \) is less than or equal to \( b_n \), or vice versa. By doing so, you can infer the behavior of the series based on the more intuitive one.
For example, if \( \sum b_n \) is a known divergent series and \( a_n \geq b_n \) for all \( n \), then \( \sum a_n \) also diverges. Conversely, if \( \sum b_n \) converges and \( a_n \leq b_n \), then \( \sum a_n \) converges as well.
This test is beneficial when deploying it alongside approximations, as seen in the problem where we simplify the behavior of \( n \sin \frac{1}{n} \). By knowing \( \sin \frac{1}{n} < \frac{1}{n} \), we conclude \( n \sin \frac{1}{n} > 1 \). Therefore, seeing \( \sum_{n=1}^{\infty} 1 \) diverges, \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \) also diverges according to the Comparison Test.
Divergent Series
A divergent series implies that when you add up the terms of the series, the sum does not settle to a finite limit. Rather, it 'diverges' to infinity or fails to approach any particular number.
The hallmark of a divergent series is that it's infinite in some sense—either because its terms do not vanish (do not get negligibly small as \( n \) approaches infinity) or because the positive and negative terms do not cancel each other out sufficiently.
A classic example is the series \( \sum_{n=1}^{\infty} 1 \), where adding an infinite number of 1's leads inevitably to infinity. This is also described in the original solution to demonstrate the divergence behavior of \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \).
Understanding which series diverge is crucial because such series often cannot be easily simplified or manipulated to prove convergence using standard tests, and they play an essential role in calculus and analysis when understanding growth rates, limits, and even functions.
The hallmark of a divergent series is that it's infinite in some sense—either because its terms do not vanish (do not get negligibly small as \( n \) approaches infinity) or because the positive and negative terms do not cancel each other out sufficiently.
A classic example is the series \( \sum_{n=1}^{\infty} 1 \), where adding an infinite number of 1's leads inevitably to infinity. This is also described in the original solution to demonstrate the divergence behavior of \( \sum_{n=1}^{\infty} n \sin \frac{1}{n} \).
Understanding which series diverge is crucial because such series often cannot be easily simplified or manipulated to prove convergence using standard tests, and they play an essential role in calculus and analysis when understanding growth rates, limits, and even functions.
Power Series
Power series are a type of infinite series involving terms of the form \( a_n x^n \), where \( a_n \) are coefficients and \( x \) is a variable. These series take the shape \( \sum_{n=0}^{\infty} a_n x^n \) and are of particular interest due to their role in functions and calculus.
Each power series has a radius of convergence, within which the series converges absolutely. Outside this range, the series can behave divergently.
Though not central to the specific exercise, understanding power series aids in recognizing convergence when variable terms are present, and they form the basis of more complex series analysis. Calculus students often engage with Taylor and Maclaurin series, which are specific types of power series used to approximate functions as sums of polynomial terms about a point.
To analyze such series, tools like the ratio test or root test often prove invaluable. These tests help determine the range of \( x \) for which the series converges, offering great insight into function behavior across intervals.
Each power series has a radius of convergence, within which the series converges absolutely. Outside this range, the series can behave divergently.
Though not central to the specific exercise, understanding power series aids in recognizing convergence when variable terms are present, and they form the basis of more complex series analysis. Calculus students often engage with Taylor and Maclaurin series, which are specific types of power series used to approximate functions as sums of polynomial terms about a point.
To analyze such series, tools like the ratio test or root test often prove invaluable. These tests help determine the range of \( x \) for which the series converges, offering great insight into function behavior across intervals.
Other exercises in this chapter
Problem 35
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