Problem 64 - Thomas Calculus

Question

In Exercises \(57 - 82 ,\) use any method to determine whether the series converges or diverges. Give reasons for your answer. $$ \sum _ { n = 2 } ^ { \infty } \frac { 3 } { 10 + n ^ { 4 / 3 } } $$

Step-by-Step Solution

Answer

The series converges by the Limit Comparison Test with a convergent p-series.

Step 1: Determine the Form of the Series

Consider the given series:\[\sum _ { n = 2 } ^ { \infty } \frac { 3 } { 10 + n ^ { 4 / 3 } }\]This is a series of positive terms. We aim to analyze the behavior of the terms \(a_n = \frac { 3 } { 10 + n ^ { 4 / 3 } }\) as \(n\) approaches infinity.

Step 2: Use the Limit Comparison Test

The Limit Comparison Test is useful here. Compare \(a_n\) with a simpler series that you know how to analyze. Consider the series \(b_n = \frac{1}{n^{4/3}}\), which is a p-series with \(p = 4/3 > 1\), thus convergent.Compute:\[\lim_{{n \to \infty}} \frac{3}{10 + n^{4/3}} \bigg/ \frac{1}{n^{4/3}} = \lim_{{n \to \infty}} \frac{3n^{4/3}}{10 + n^{4/3}} \]Simplify the expression:\[\lim_{{n \to \infty}} \frac{3n^{4/3}}{n^{4/3}(\frac{10}{n^{4/3}} + 1)} = \lim_{{n \to \infty}} \frac{3}{\frac{10}{n^{4/3}} + 1} = 3\]Since the limit is 3, which is a finite positive number, both series \(\sum a_n\) and \(\sum b_n\) either converge or diverge together.

Step 3: Conclusion Upon Comparison

Since \(\sum b_n = \sum \frac{1}{n^{4/3}}\) is a convergent p-series (as p > 1), the given series \(\sum a_n = \sum \frac{3}{10 + n^{4/3}}\) also converges by the Limit Comparison Test.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a powerful tool used to determine the convergence or divergence of series. It provides a way to compare a complex series to a simpler one that we already understand well. To use this test, you select a series, let's call it \( b_n \), that is similar in behavior to your series \( a_n \). Here's how the test works:

• Compute the limit \( L = \lim_{{n \to \infty}} \frac{a_n}{b_n} \).
• If \( L \) is a positive, finite number, then both series \( \sum a_n \) and \( \sum b_n \) will either both converge or both diverge.

For example, in this exercise, we derived \( a_n = \frac{3}{10 + n^{4/3}} \) and used \( b_n = \frac{1}{n^{4/3}} \) as the comparison series. The limit computed was a finite positive number (3), confirming that both series behave identically regarding convergence.

P-Series

A p-series is a specific type of series that takes the form \( \sum \frac{1}{n^p} \). The convergence or divergence of a p-series depends on the value of \( p \):

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

In the context of the exercise, we compared the given series to the p-series \( \sum \frac{1}{n^{4/3}} \). This particular series converges because \( p = \frac{4}{3} \) which is greater than 1. Recognizing and using p-series allows us to draw quick and powerful conclusions about other series by comparison.

Positive Terms

Series with positive terms exhibit simpler convergence properties that make them easier to analyze. When all terms of a series are positive, the series either converges to a specific value or diverges to infinity.Positive term series do not oscillate, which means they steadily move in one direction, either increasing or stabilizing. This is particularly useful when applying the Limit Comparison Test because positivity ensures that if \( a_n \) is bounded by another convergent series, \( b_n \), then \( a_n \) will also converge. Conversely, if \( b_n \) diverges, \( a_n \) will also diverge.

Asymptotic Behavior

Asymptotic behavior refers to how a function or sequence behaves as its input goes to infinity. For series, understanding asymptotic behavior helps simplify the terms involved, especially when the expression looks complex.In the given series, \( a_n = \frac{3}{10 + n^{4/3}} \), for large \( n \), the term \( n^{4/3} \) dominates the constant 10. Therefore, \( a_n \) behaves similarly to the simpler term \( \frac{3}{n^{4/3}} \) as \( n \to \infty \). Recognizing this dominant term helps choose the best series for the Limit Comparison Test.With asymptotic behavior, simplification provides clarity in analysis, which ultimately leads to correctly determining the convergence or divergence of complex series.