Problem 48 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{\sec ^{-1} n}{n^{1.3}} $$

Step-by-Step Solution

Answer

The series diverges.

Step 1: Identify the General Term

The given series is \( \sum_{n=1}^{\infty} \frac{\sec^{-1} n}{n^{1.3}} \). This series has the general term \( a_n = \frac{\sec^{-1} n}{n^{1.3}} \). Our task is to determine the convergence or divergence of this series.

Step 2: Consider the Limit Comparison Test (LCT)

To analyze the convergence, we use the limit comparison test with a simpler series, \( \sum_{n=1}^{\infty} \frac{1}{n^{1.3}} \), which is a p-series with \( p = 1.3\). We know this p-series converges because \( p > 1 \).

Step 3: Apply the Limit Comparison Test

Compute the limit \( \lim_{n \to \infty} \frac{\frac{\sec^{-1} n}{n^{1.3}}}{\frac{1}{n^{1.3}}} = \lim_{n \to \infty} \sec^{-1} n \). The function \( \sec^{-1} n \) is asymptotically equivalent to \( \ln n \) for large values of \( n \), so the limit diverges to infinity.

Step 4: Interpret the Limit

Since the limit diverges to infinity, the limit comparison test is inconclusive. We must try another method to determine convergence or divergence.

Step 5: Consider the Divergence Test

The divergence test states that if \( \lim_{n \to \infty} a_n eq 0 \), then the series diverges. For our series, \( a_n = \frac{\sec^{-1} n}{n^{1.3}} \). As \( n \to \infty \), \( \sec^{-1} n \to \infty \), so \( a_n \to \infty \). Therefore, \( \lim_{n \to \infty} a_n eq 0 \), indicating divergence.

Step 6: Conclusion of Divergence

Since \( a_n \) does not tend to zero as \( n \to \infty \), the series \( \sum_{n=1}^{\infty} \frac{\sec^{-1} n}{n^{1.3}} \) diverges by the divergence test.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a powerful tool for determining whether a series converges or diverges. In this approach, you compare a given series \( \sum a_n \) with another series \( \sum b_n \) that is simpler and whose behavior you already understand. For the Limit Comparison Test to work:

• The series you are comparing must have positive terms.
• You then calculate the limit \ \lim_{n \to \infty} \frac{a_n}{b_n} \, which should be a finite positive number.

If this limit is finite and positive, and the series \( \sum b_n \) is known to converge or diverge, the behavior of \( \sum a_n \) will be the same. In the original exercise, \( \lim_{n \to \infty} \sec^{-1} n = \infty \) shows the Limit Comparison Test is inconclusive because the limit does not provide a finite positive result. This suggests exploring another test for further analysis.

Divergence Test

The Divergence Test is straightforward but crucial for testing series convergence. It states that if \( \lim_{n \to \infty} a_n eq 0 \), then the series \( \sum a_n \) must diverge. In simple terms, for a series to have any chance of converging, the general term \( a_n \) must approach zero as \( n \) becomes infinitely large.In the given series \( \sum_{n=1}^{\infty} \frac{\sec^{-1} n}{n^{1.3}} \), we find that as \( n \) grows, \( \sec^{-1} n \) increases without bound. Consequently, \( a_n \) does not tend towards zero, making the divergence test a decisive way to conclude that the series diverges. It's a powerful initial check before deeper evaluations needing intricate analyses.

P-Series

A P-Series is a specific type of series that takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The convergence of a p-series depends on the value of \( p \):

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

In the provided solution, the comparison series \( \sum_{n=1}^{\infty} \frac{1}{n^{1.3}} \) is a p-series with \( p = 1.3 \). Since \( 1.3 > 1 \), this series is convergent. This series was used in the limit comparison test as a benchmark to initially investigate the nature of the more complicated original series, although it did not conclude the analysis.

General Term

The general term in a series, typically denoted as \( a_n \), is crucial as it dictates the behavior of the series for large \( n \). It represents the \( n^{th} \) term of the sequence being summed.For the series \( \sum_{n=1}^{\infty} \frac{\sec^{-1} n}{n^{1.3}} \), the general term is \( a_n = \frac{\sec^{-1} n}{n^{1.3}} \). Understanding the growth pattern of \( a_n \) is essential to testing series convergence.In this case, despite \( n^{1.3} \) growing in the denominator, \( \sec^{-1} n \) increases rapidly enough for \( a_n \) not to approach zero. This insight, particularly when \( a_n ot\to 0 \), leads directly to applying the divergence test, confirming that the series diverges.