Problem 13 - Thomas Calculus

Question

In Exercises \(9-16,\) use the Root Test to determine if each series converges absolutely or diverges. $$\sum_{n=1}^{\infty} \frac{-8}{(3+(1 / n))^{2 n}}$$

Step-by-Step Solution

Answer

The series converges absolutely by the Root Test.

Step 1: Identify the Series

The given series is \( \sum_{n=1}^{\infty} \frac{-8}{(3+(1 / n))^{2 n}} \). To apply the Root Test, we need to consider the general term \( a_n = \frac{-8}{(3+(1 / n))^{2 n}} \).

Step 2: Apply the Root Test

The Root Test involves finding \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \). Here, \(|a_n| = \frac{8}{(3+(1/n))^{2n}}\). Thus, we need to evaluate \( \lim_{n \to \infty} \sqrt[n]{\frac{8}{(3+(1/n))^{2n}}} \).

Step 3: Simplify the Expression

Simplify the expression under the root: \( \sqrt[n]{\frac{8}{(3+(1/n))^{2n}}} = \frac{\sqrt[n]{8}}{3+(1/n)} \). As \(n\) grows large, \(1/n\) approaches zero, thus \(3+(1/n) \) approaches 3. Also, \(\sqrt[n]{8}\) approaches 1, because the nth root of a constant approaches 1 as n goes to infinity.

Step 4: Evaluate the Limit

Now we determine the limit: \( \lim_{n \to \infty} \frac{1}{3} = \frac{1}{3} \).

Step 5: Conclusion Based on Root Test

Since the result of the limit is \( \frac{1}{3} < 1 \), by the Root Test, the given series converges absolutely.

Key Concepts

Convergent Series

A **convergent series** is one where the sum of its infinite terms approaches a specific finite number. We know this is true if the sum gets closer and closer to a number as we keep adding terms. Convergent series are important in mathematical analysis because they help us understand how functions behave when expressed as infinite sums. For instance:

• Mathematicians use convergent series to solve complex mathematical equations and problems.
• They allow us to approximate values which might be difficult to compute directly.

In the exercise, we are asked to determine if a given series is convergent by employing the Root Test. If the Root Test indicates a value less than 1, it means the series converges absolutely and, by default, it also converges. This gives a robust method to quickly identify convergence without manually calculating all sums.

Absolute Convergence

**Absolute convergence** of a series happens when the series is convergent even after replacing every term with its absolute value. This concept simplifies many calculations and theoretical checks in mathematics. In particular:

• If a series is absolutely convergent, it is also convergent by default.
• Absolute convergence implies the series behaves predictably regardless of the arrangement of its terms, making it very stable and reliable in numerical analysis.

In the provided solution, the Root Test was employed to confirm absolute convergence. Since \( \frac{1}{3} \) is less than 1, the result signals that the original series, which included negative terms due to the factor of \( -8 \), converges absolutely. This means that not only does the sequence of sums approach a finite number, but the arrangement of its terms does not affect this outcome.

Limit Evaluation

Evaluating limits accurately is crucial in applying the Root Test. In our exercise, the Root Test involves assessing the limit:\[ \lim_{n \to \infty} \sqrt[n]{|a_n|} \]where \( |a_n| \) is the absolute value of the series' general term. This limits all values that could become infinitely large or small, allowing for a concise result that tells us about the series' behavior.

• The Root Test outcome depends heavily on understanding how limits work as \( n \) tends to infinity.
• Typically involves simplifying expressions—like reducing \( 1/n \) terms so they have no effect as \( n \) becomes large.

In our task, simplification through limit evaluation allowed us to conclude that \( \lim_{n \to \infty} \sqrt[n]{8} / (3 + 1/n) \) approaches \( 1/3 \). Recognizing that terms like \( 1/n \) shrink to zero was key to understanding the ultimate behavior of the series. This step-by-step simplification and limit evaluation perfectly illustrate the method's power and simplicity.