Problem 62 - 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 \(n^2 a_n \to 0\), making \(a_n\) smaller than the convergent series \(\frac{1}{n^2}\) for large \(n\).

Step 1: Understand the Problem

We are given a sequence \(a_n > 0\) and know that \(\lim _{n \rightarrow \infty} n^{2} a_{n}=0\). We are asked to prove that the series \(\sum a_{n}\) converges.

Step 2: Applying the Limit

Since \(\lim _{n \rightarrow \infty} n^{2} a_{n}=0\), it means that as \(n\) becomes very large, \(n^{2} a_{n}\) approaches 0. This indicates that \(a_{n}\) must be approaching 0, and specifically, it does so faster than \(\frac{1}{n^2}\) grows.

Step 3: Using a Comparison Test

To prove the convergence of \(\sum a_n\), consider comparing it with a known convergent series. The sequence \(b_n = \frac{1}{n^2}\) converges (p-series with \(p > 1\)). Since \(n^{2} a_{n}\) approaches 0, \(a_n\) is eventually smaller than \(\frac{1}{n^2}\) for sufficiently large \(n\).

Step 4: Constructing the Comparison

For large \(n\), let's say \(a_n < C \cdot \frac{1}{n^2}\) for some constant \(C\) (since \(n^2 a_n \rightarrow 0\)). Here, \(C\) can be any constant such that this inequality holds for sufficiently large \(n\).

Step 5: Applying the Limit Comparison Test

The Limit Comparison Test states that if \(a_n > 0\), \(b_n > 0\), and \(\lim_{n \to \infty} \frac{a_n}{b_n} = L\) where \(0 < L < \infty\), then \(\sum a_n\) converges if and only if \(\sum b_n\) converges. Here, \(\frac{a_n}{b_n} = a_n n^2\) which approaches 0. Thus, for large enough \(n\), even though \(L\) might not fit the regular condition exactly, intuitively, \(a_n\) is dominated by \(b_n\).

Step 6: Conclusion

Since \(a_n\) is eventually smaller than \(\frac{1}{n^2}\) and \(\sum \frac{1}{n^2}\) converges, \(\sum a_n\) must also converge by the comparison or limit comparison test.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a mathematical tool used to determine the convergence of infinite series. It is especially useful when comparing a given series to a known benchmark series with known behavior, such as a p-series.
The test essentially says: If you have two positive-term series, \ \sum a_{n} \ and \ \sum b_{n} \, compute \(\lim_{n \to \infty} \frac{a_{n}}{b_{n}} = L\) and if \(0 < L < \infty\), then both series converge or diverge together.
When applying this test, select a comparison series \ b_{n} \ whose convergence behavior is already known. In our case, \ b_{n} = \frac{1}{n^2} \ is a common choice because it represents a convergent series (since \ p > 1 \ in the p-series). Through this test, if the limit of the ratio is finite and positive, we can infer that the behavior of \ \sum a_{n} \ is similar to that of \ \sum b_{n} \, leading us to the conclusion about the convergence of \ \sum a_{n} \.

P-Series

A p-series is a type of series in the form \ \sum \frac{1}{n^p} \, where \(n\) ranges from 1 to infinity. This series is known as a benchmark in analyzing convergence.
The convergence of a p-series is determined by the value of \(p\):

• If \(p > 1\), the series converges.
• If \(p \leq 1\), the series diverges.

In this exercise, the series we use for comparison is \(\sum \frac{1}{n^2}\), which is a p-series with \(p = 2\). Since \(p = 2 > 1\), this series converges. By using p-series, we often benchmark the convergence or divergence behavior of more complicated series.

Sequence and Series

A sequence is a set of numbers organized in a specific order. These numbers are indexed by natural numbers, like \(a_{1}, a_{2}, a_{3}, \ldots\). A series, on the other hand, involves summing the terms of a sequence, represented as \(\sum a_{n}\).
The convergence of a series refers to whether this sum approaches a finite value as the number of terms in the sequence increases indefinitely. If the sum of the sequence terms approaches a limit as the number of terms tends towards infinity, we say the series converges.
In our given problem, the series in question is \(\sum a_{n}\), and we are tasked with showing it converges by comparing it to another series, \(\sum \frac{1}{n^2}\). Sequences and series are foundational topics in calculus and analysis, where understanding how individual sequence terms partakes in the broader context of their sum is critical.

Limit of a Sequence

Limits of sequences address the behavior of sequence terms as the index \(n\) becomes increasingly large. Specifically, it tells us how terms of a sequence \(a_{n}\) behave when \(n\) goes towards infinity. This concept is crucial when dealing with series.

• If \(\lim_{n \to \infty} a_{n} = L\) and \(L\) is finite, the terms of the sequence are approaching a specific fixed value.
• If \(L = 0\), this could indicate that the associated series \(\sum a_{n}\) might converge, provided other conditions (like the rate of decrease) are satisfied.

In the given exercise, we know that \(\lim _{n \rightarrow \infty} n^{2} a_{n}=0\). This reveals that the terms \(a_{n}\) decrease to zero faster than \(\frac{1}{n^2}\) increases, suggesting convergence of the series. The limit of a sequence gives essential insight into determining convergence by understanding the tail behavior of the series.