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Suppose that \(\lim _{n \rightarrow \infty} n a_{n}=1\). Prove that \(\Sigma a_{n}\) diverges.

Short Answer

Expert verified
The series \( \Sigma a_n \) diverges because \( a_n \approx \frac{1}{n} \), similar to the divergent harmonic series.

Step by step solution

01

Understanding the Problem

We are given that \( \lim_{n \rightarrow \infty} n a_n = 1 \). We need to prove that the series \( \Sigma a_n \) diverges.
02

Rewriting the Limit Statement

The statement \( \lim_{n \rightarrow \infty} n a_n = 1 \) means that as \( n \) becomes very large, the expression \( n a_n \) gets arbitrarily close to 1. Thus, for sufficiently large \( n \), we have that \( n a_n \approx 1 \).
03

Understanding Implications on \( a_n \)

Since \( n a_n \approx 1 \), it follows that \( a_n \approx \frac{1}{n} \) for large \( n \). This means that the terms \( a_n \) behave similarly to the harmonic series terms.
04

Connection to Harmonic Series

The harmonic series \( \Sigma \frac{1}{n} \) is a well-known divergent series. If \( a_n \approx \frac{1}{n} \), then the series \( \Sigma a_n \) behaves similarly to the harmonic series, suggesting divergence.
05

Divergence Conclusion

Since \( n a_n \) approaches 1 as \( n \) increases, and \( a_n \approx \frac{1}{n} \) mimicking the harmonic series' behavior, we conclude that \( \Sigma a_n \) must also diverge, because it lacks the property to reach a finite limit.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit of a sequence
The limit of a sequence is a fundamental concept in calculus and analysis. When we speak of the limit of a sequence, we consider what happens to the sequence's terms as the sequence progresses infinitely. In the given problem, we have the sequence term expressed as \( n a_n \), and we are examining its behavior as \( n \) approaches infinity. Here, \( \lim_{n \rightarrow \infty} n a_n = 1 \) tells us that as we look at larger and larger values of \( n \), the product \( n a_n \) becomes increasingly close to 1. This suggests that multiplying \( a_n \) by larger \( n \) will approximate 1 more closely.

To delve deeper, this implies that \( a_n \) alone must shrink in size with increasing \( n \), such that when multiplied by \( n \), it remains near 1. Specifically, if \( n a_n \approx 1 \), then \( a_n \approx \frac{1}{n} \). Understanding this behavior helps us relate sequence limits to series behavior, guiding us on whether the associated series might converge or diverge.
Harmonic series
The harmonic series is a classic example of a divergent infinite series. It is defined as \( \Sigma \frac{1}{n} \) with \( n \) starting from 1 and going to infinity. Despite the terms \( \frac{1}{n} \) decreasing and approaching zero, the series as a whole does not converge to a finite sum. This is a well-established fact in mathematics, which makes it a standard point of reference for evaluating other similar series.

In our exercise, we demonstrated that \( a_n \approx \frac{1}{n} \). Thus, the series \( \Sigma a_n \) behaves similarly to the harmonic series. Since the harmonic series itself diverges, we are led to conclude that any series which closely parallels the structure of a harmonic series must also diverge. This connection is pivotal, as it provides a clear and often simpler way to evaluate series with terms that mimic these characteristics.
Comparison test for series
The comparison test for series is a powerful tool in determining whether a series converges or diverges. The basic idea is to compare the series in question with another series whose convergence behavior is already known. This is a useful technique because it allows us to apply the known properties of simpler series, like the harmonic series, to more complex scenarios.

In this case, observing that \( a_n \approx \frac{1}{n} \), we compare our series \( \Sigma a_n \) with the harmonic series, \( \Sigma \frac{1}{n} \), which is known to diverge. By aligning \( a_n \) closely with \( \frac{1}{n} \), we assert through the comparison test that if \( \Sigma \frac{1}{n} \) diverges, then \( \Sigma a_n \) must also diverge. This test relies on the condition that our series terms are larger than, equal to, or closely mimic those of a known divergent series over the long term, reinforcing the conclusion of divergence.

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