Problem 60 - Thomas Calculus

Question

Suppose that \(a_{n}>0\) and \(\lim _{n \rightarrow \infty} n^{2} a_{n}=0 .\) Prove that \(\sum a_{n}\) converges.

Step-by-Step Solution

Answer

The series \( \sum a_n \) converges because \( a_n \) decays faster than \( \frac{1}{n^2} \).

Step 1: Understanding the Limit Condition

Given that \( \lim_{n \to \infty} n^2 a_n = 0 \), we know that for sufficiently large \( n \), the terms \( n^2 a_n \) become arbitrarily small. This suggests that \( a_n \) must decay faster than \( \frac{1}{n^2} \) as \( n \) becomes very large.

Step 2: Establishing Asymptotic Behavior

For very large \( n \), since \( n^2 a_n \to 0 \), the term \( a_n \) must behave like \( \frac{1}{n^{2+\epsilon}} \) for some \( \epsilon > 0 \). This is because if \( a_n \sim \frac{1}{n^2} \), it contradicts the fact that \( n^2 a_n \to 0 \).

Step 3: Comparing with a Known Convergent Series

Consider the series \( \sum \frac{1}{n^{2+\epsilon}} \). Since \( 2+\epsilon > 1 \), according to the \(\textit{p-series}\) test, the series \( \sum \frac{1}{n^{2+\epsilon}} \) converges.

Step 4: Applying the Comparison Test

Since \( a_n \sim \frac{1}{n^{2+\epsilon}} \) for large \( n \), there exists some constant \( C > 0 \) such that for all sufficiently large \( n \), \( a_n \leq C \cdot \frac{1}{n^{2+\epsilon}} \). By the comparison test, since \( \sum \frac{1}{n^{2+\epsilon}} \) converges, \( \sum a_n \) also converges.

Key Concepts

p-series test

The p-series test is an important tool for determining the convergence of an infinite series. This test specifically applies to series of the form \( \sum \frac{1}{n^p} \) where \( p \) is a positive constant.
Simply put, a p-series converges if \( p > 1 \) and diverges if \( p \leq 1 \). This is a fundamental rule to remember, as it helps in rapidly evaluating the behavior of many series without complicated calculations.

• For example, when \( p = 2 \), the series \( \sum \frac{1}{n^2} \) converges, meaning it sums to a finite value as \( n \) approaches infinity.
• In contrast, if \( p = 1 \), the series \( \sum \frac{1}{n} \) diverges, suggesting its sum grows without bound.

In the original problem, a p-series came into play as we considered \( \sum \frac{1}{n^{2+\epsilon}} \) for some small positive \( \epsilon \). Since \( 2 + \epsilon > 1 \), we can confidently use the p-series test to assert convergence.
Understanding this concept is crucial when paired with other techniques like the comparison test for a comprehensive approach to solving series problems.

Comparison Test

The comparison test is a handy method used to determine convergence by comparing an unknown series with a known benchmark. It is particularly useful when direct methods are complex or unavailable.
The basic idea is straightforward:

• If a series \( \sum a_n \) is such that \( 0 \leq a_n \leq b_n \) for all \( n \) past some point, and \( \sum b_n \) is known to converge, then \( \sum a_n \) must also converge.
• Conversely, if \( \sum b_n \) diverges and \( a_n \geq b_n \) for sufficiently large \( n \), then \( \sum a_n \) also diverges.

The test is a powerful approach when dealing with challenging series. It allows you to leverage known results. In our exercise, the series \( \sum \frac{1}{n^{2+\epsilon}} \) converges according to the p-series test. Since \( a_n \leq C \cdot \frac{1}{n^{2+\epsilon}} \), by the comparison test, we can safely conclude \( \sum a_n \) converges.

Asymptotic Behavior

Asymptotic behavior in mathematics refers to the tendency of a function as an input approaches a boundary or infinity. It's a way to describe how a series behaves by comparing it to a simpler function or sequence.To apply this concept, consider \( n^2 a_n \to 0 \). This indicates that as \( n \) becomes larger and larger, \( a_n \) behaves like \( \frac{1}{n^{2 + \epsilon}} \) for some \( \epsilon > 0 \). This is an asymptotic comparison, where the leading term is dominant, and the rest of the terms are negligible.

• This helps to identify the general trend of decay in \( a_n \) as \( n \to \infty \).
• Asymptotic behavior simplifies complex expressions to a form that is easier to analyze and compare with other functions or series.

In our problem, understanding that \( a_n \sim \frac{1}{n^{2+\epsilon}} \) helped us apply the comparison test effectively. Asymptotic analysis was key in approximating \( a_n \), indicating that it decays faster than \( \frac{1}{n^2} \), thereby proving the convergence of the series \( \sum a_n \).