Problem 39
Question
Suppose that \(\lim _{n \rightarrow \infty} n a_{n}=1 .\) Prove that \(\sum a_{n}\) diverges.
Step-by-Step Solution
Verified Answer
The series \( \sum a_n \) diverges because \( a_n \approx \frac{1}{n} \) and the harmonic series diverges.
1Step 1: Understanding the Limit
Given that \( \lim _{n \rightarrow \infty} n a_{n} = 1 \), this means that as \( n \) becomes very large, the product \( n a_{n} \) approaches 1. Essentially, as \( n \) increases, the sequence \( a_n \) behaves like \( \frac{1}{n} \).
2Step 2: Exploring Sequence Behavior
If \( n a_n \) tends to 1, then \( a_n \) is approximately \( \frac{1}{n} \) for very large \( n \). For example, if \( a_n = \frac{1}{n} + \frac{1}{n^2} \), then the sequence behaves similarly, where the dominant part is \( \frac{1}{n} \).
3Step 3: Applying the Harmonic Series Divergence
The harmonic series \( \sum \frac{1}{n} \) is known to diverge. Since \( a_n \approx \frac{1}{n} \), the series \( \sum a_{n} \) will similarly diverge like the harmonic series.
4Step 4: Concluding with Comparison Test
Using the comparison test, which states that if \( a_n \approx \frac{1}{n} \) and \( \sum \frac{1}{n} \) diverges, then \( \sum a_n \) must also diverge. Thus, \( \sum a_n \) diverges.
Key Concepts
Limit of SequencesHarmonic SeriesComparison Test
Limit of Sequences
An essential concept in understanding divergent series is the limit of sequences. A sequence is a list of numbers in a specific order, and when we speak about the limit of a sequence as \( n \) approaches infinity, we're interested in the value that the sequence approaches as \( n \) becomes very large.
In this exercise, we have a sequence \( a_n \) such that \( \lim _{n \rightarrow \infty} n a_{n} = 1 \). This implies that as \( n \) grows, the product \( n a_{n} \) gets closer and closer to 1. Therefore, for large \( n \), \( a_n \) behaves roughly like \( \frac{1}{n} \). This is a key observation because it helps us compare \( a_n \) to known sequences, like the reciprocal function \( \frac{1}{n} \).
The behavior of sequences as they approach their limits forms the cornerstone of many mathematical proofs and is crucial in determining the convergence or divergence of series.
In this exercise, we have a sequence \( a_n \) such that \( \lim _{n \rightarrow \infty} n a_{n} = 1 \). This implies that as \( n \) grows, the product \( n a_{n} \) gets closer and closer to 1. Therefore, for large \( n \), \( a_n \) behaves roughly like \( \frac{1}{n} \). This is a key observation because it helps us compare \( a_n \) to known sequences, like the reciprocal function \( \frac{1}{n} \).
The behavior of sequences as they approach their limits forms the cornerstone of many mathematical proofs and is crucial in determining the convergence or divergence of series.
Harmonic Series
The harmonic series is a classic example of a divergent series. It is represented as the sum \( \sum \frac{1}{n} \), starting at \( n = 1 \). Even though the terms \( \frac{1}{n} \) become smaller and smaller, the series does not converge to a finite number; instead, it diverges.
Understanding that the harmonic series diverges is a critical piece of knowledge when analyzing other series. The series in question in the exercise, \( \sum a_n \), has terms \( a_n \) that behave like \( \frac{1}{n} \) for large \( n \). The comparison to the harmonic series becomes evident, indicating a similar divergent nature.
Understanding that the harmonic series diverges is a critical piece of knowledge when analyzing other series. The series in question in the exercise, \( \sum a_n \), has terms \( a_n \) that behave like \( \frac{1}{n} \) for large \( n \). The comparison to the harmonic series becomes evident, indicating a similar divergent nature.
- Key Point: The harmonic series \( \sum \frac{1}{n} \) is a benchmark for testing the divergence of series with comparable terms.
Comparison Test
The comparison test is a powerful tool in calculus for determining the convergence or divergence of infinite series. The essence of this test is simple: if you have a series with terms \( a_n \) and you know another series with terms \( b_n \) that is easier to analyze, you can compare them to draw conclusions.
In practice, if every term \( a_n \) is less than or equal to the corresponding term \( b_n \), and \( \sum b_n \) converges, then \( \sum a_n \) also converges. Conversely, if \( b_n \) is smaller and its series diverges, as is the case in the exercise, then \( \sum a_n \) also diverges.
For our exercise, since \( a_n \approx \frac{1}{n} \) and we know \( \sum \frac{1}{n} \) diverges, the comparison test allows us to confidently assert that \( \sum a_n \) also diverges.
In practice, if every term \( a_n \) is less than or equal to the corresponding term \( b_n \), and \( \sum b_n \) converges, then \( \sum a_n \) also converges. Conversely, if \( b_n \) is smaller and its series diverges, as is the case in the exercise, then \( \sum a_n \) also diverges.
For our exercise, since \( a_n \approx \frac{1}{n} \) and we know \( \sum \frac{1}{n} \) diverges, the comparison test allows us to confidently assert that \( \sum a_n \) also diverges.
- The comparison test uses known behaviors of standard series to draw conclusions about similar series.
Other exercises in this chapter
Problem 38
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