Problem 36
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.) $$ \sum_{n=1}^{\infty} n \tan \frac{1}{n} $$
Step-by-Step Solution
Verified Answer
The series \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \) diverges by the Limit Comparison Test.
1Step 1: Identify the series and the test to use
The given series is \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \). To determine its convergence, we can use the Limit Comparison Test. For this test, we find a series \( \sum_{n=1}^{\infty} b_n \) such that \( b_n \) behaves similarly as \( a_n = n \tan \frac{1}{n} \) for large \( n \). We will use \( b_n = n \frac{1}{n} = 1 \).
2Step 2: Simplify the expression for large n
As \( n \to \infty \), the expression \( \tan \frac{1}{n} \approx \frac{1}{n} \), since \( \tan x \approx x \) when \( x \) is small. So, \( n \tan \frac{1}{n} \approx n \cdot \frac{1}{n} = 1 \). Thus, \( a_n \approx 1 \) for large \( n \).
3Step 3: Apply Limit Comparison Test
The Limit Comparison Test states that if \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \) where \( 0 < c < \infty \), then \( \sum_{n=1}^{\infty} a_n \) and \( \sum_{n=1}^{\infty} b_n \) both converge or both diverge. Here, \( b_n = 1 \). We compute:\[\lim_{n \to \infty} \frac{n \tan \frac{1}{n}}{1} = \lim_{n \to \infty} n \tan \frac{1}{n} = \lim_{n \to \infty} 1 = 1.\]Since \( b_n = 1 \) leads to the harmonic series \( \sum_{n=1}^{\infty} 1 \), which diverges, \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \) also diverges.
Key Concepts
Limit Comparison TestDivergence of SeriesHarmonic Series
Limit Comparison Test
The limit comparison test is a handy tool for determining the convergence or divergence of an infinite series. It's particularly useful when direct comparison isn't straightforward. The principle behind this test is to compare your series to another series whose behavior you already know.
The idea is to find another series, called the "comparison series", which behaves similarly to your original series. Let's say your series is \( \sum a_n \) and your comparison series is \( \sum b_n \). You calculate the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \).
For example, in the given exercise, by choosing \( b_n = n \frac{1}{n} = 1 \), we leveraged the known divergence of the harmonic series \( \sum \frac{1}{n} \) to show that \( \sum n \tan \frac{1}{n} \) also diverges.
The idea is to find another series, called the "comparison series", which behaves similarly to your original series. Let's say your series is \( \sum a_n \) and your comparison series is \( \sum b_n \). You calculate the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \).
- If \( 0 < c < \infty \), both series either converge or they both diverge.
- If \( c = 0 \) or \( c = \infty \), the test is inconclusive, and you must try another method.
For example, in the given exercise, by choosing \( b_n = n \frac{1}{n} = 1 \), we leveraged the known divergence of the harmonic series \( \sum \frac{1}{n} \) to show that \( \sum n \tan \frac{1}{n} \) also diverges.
Divergence of Series
Understanding the divergence of a series is fundamental in calculus. A series diverges if the sum of its terms does not settle down to a fixed number. In simpler terms, a divergent series keeps growing and doesn't approach a limit as more terms are added.
When using convergence tests, like the limit comparison test, identifying divergence is often a key outcome. If you find that your series relates to a divergent series through a valid limit comparison, you immediately conclude that your series diverges too. This process simplifies the analysis of complicated series.
For the series \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \), the comparison with the harmonic series shows divergence. Here, the harmonic series \( \sum_{n=1}^{\infty} 1 \) diverges. Thus, by finding that \( n \tan \frac{1}{n} \) behaves similarly to 1 for large \( n \), it's demonstrated that it diverges as well.
Recognizing when and why a series diverges is crucial for understanding the behavior of infinite sums in mathematics.
When using convergence tests, like the limit comparison test, identifying divergence is often a key outcome. If you find that your series relates to a divergent series through a valid limit comparison, you immediately conclude that your series diverges too. This process simplifies the analysis of complicated series.
For the series \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \), the comparison with the harmonic series shows divergence. Here, the harmonic series \( \sum_{n=1}^{\infty} 1 \) diverges. Thus, by finding that \( n \tan \frac{1}{n} \) behaves similarly to 1 for large \( n \), it's demonstrated that it diverges as well.
Recognizing when and why a series diverges is crucial for understanding the behavior of infinite sums in mathematics.
Harmonic Series
The harmonic series is one of the most famous examples of a divergent series. It takes the form \( \sum_{n=1}^{\infty} \frac{1}{n} \), and despite its terms getting smaller and smaller, the sum of these terms grows indefinitely without bound.
In calculus, it serves as a benchmark for divergence. If you can show that another series has terms that are comparable to \( \frac{1}{n} \) for large \( n \), you can often conclude that your series diverges as well. This makes understanding the harmonic series essential for any student working with series convergence tests.
In the given problem, comparing the complicated series \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \) to the simple harmonic series was pivotal. By finding that \( n \tan \frac{1}{n} \approx 1 \) for large \( n \), and knowing \( \sum 1 \) diverges, we conclude the original series diverges too.
This classic series is a powerful example of how deep mathematical concepts can revolve around simple principles.
In calculus, it serves as a benchmark for divergence. If you can show that another series has terms that are comparable to \( \frac{1}{n} \) for large \( n \), you can often conclude that your series diverges as well. This makes understanding the harmonic series essential for any student working with series convergence tests.
In the given problem, comparing the complicated series \( \sum_{n=1}^{\infty} n \tan \frac{1}{n} \) to the simple harmonic series was pivotal. By finding that \( n \tan \frac{1}{n} \approx 1 \) for large \( n \), and knowing \( \sum 1 \) diverges, we conclude the original series diverges too.
This classic series is a powerful example of how deep mathematical concepts can revolve around simple principles.
Other exercises in this chapter
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