Problem 37 - Calculus: Early Transcendentals

Question

Test the series for convergence or divergence. \( \displaystyle \sum_{n = 1}^{\infty} (\sqrt[n]{2} - 1)^n \)

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} (\sqrt[n]{2} - 1)^n \) converges by the Root Test.

Step 1: Identify the series for convergence test

We have the series \( \sum_{n = 1}^{\infty} (\sqrt[n]{2} - 1)^n \). We need to determine whether this series converges or diverges.

Step 2: Simplify the term

Observe that \( (\sqrt[n]{2})^n = 2 \), therefore, \( \sqrt[n]{2} \approx 1 + \frac{\ln(2)}{n} \) for large \( n \). Thus, the term \( (\sqrt[n]{2} - 1) \approx \frac{\ln(2)}{n} \). Substitute this approximation into the series term to simplify it.

Step 3: Rewrite the series using approximation

Substitute the approximation \( (\sqrt[n]{2} - 1) \approx \frac{\ln(2)}{n} \) to rewrite the series as \( \sum_{n=1}^{\infty} \left(\frac{\ln(2)}{n}\right)^n \).

Step 4: Analyze the series by comparing with a p-series

Observe that the series \( \left(\frac{\ln(2)}{n}\right)^n \) resembles a general term form \( a^n \) where \( a = \frac{\ln(2)}{n} \). For large \( n \), \( \frac{\ln(2)}{n} < 1 \), leading the terms to approach zero rapidly as \( n \to \infty \). Use the comparison test to analyze this.

Step 5: Apply the Root Test

Apply the Root Test: calculate \( \lim_{n \to \infty} \sqrt[n]{|(\frac{\ln(2)}{n})^n|} = \lim_{n \to \infty} \left|\frac{\ln(2)}{n}\right| = 0 \). Since this limit is less than 1, by the Root Test, the series converges.

Key Concepts

Root Test

The Root Test is an effective way to determine whether an infinite series converges or diverges. It involves taking the limit as the nth-root of the absolute value of the series term approaches infinity. The Root Test states that for a series \( \sum a_n \), if \( \lim_{n \to \infty} \sqrt[n]{|a_n|} = L \), then:
- The series converges absolutely if \( L < 1 \).- The series diverges if \( L > 1 \).- The test is inconclusive if \( L = 1 \).

In this exercise, we applied the Root Test to the series \( \sum_{n=1}^{\infty} (\frac{\ln(2)}{n})^n \). By computing the limit, we found that it is 0, which is less than 1, indicating convergence.

p-series

A p-series is a specific type of series that takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a positive constant. The convergence of a p-series depends on the value of \( p \):
- It converges if \( p > 1 \).- It diverges if \( p \leq 1 \).

In the solution, while the series doesn't exactly fit a p-series, the approximation \( (\sqrt[n]{2} - 1) \approx \frac{\ln(2)}{n} \) resembles the harmonic series (a p-series with \( p = 1 \)), which diverges. However, since we transformed the series and applied the Root Test, we confirmed its convergence.

Limit Comparison Test

The Limit Comparison Test is a powerful tool for analyzing convergence by comparing the terms of a given series with those of a known benchmark series. It involves:

• Choosing a comparison series \( \sum b_n \) that is known to converge or diverge.
• Calculating \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \) where \( 0 < c < \infty \).
• If the limit exists and is positive, both series converge or diverge together.

In this exercise, while the Limit Comparison Test was not explicitly used, the approximation helped us keep the terms manageable for applying the Root Test,

Infinite Series

Infinite series are the sum of infinitely many terms, expressed as \( \sum_{n=1}^{\infty} a_n \). Convergence in an infinite series means the sum trends towards a finite value as more terms are added.

The behavior of each term in the context of all previous terms determines the overall behavior of the series. In this exercise, the series \( \sum_{n = 1}^{\infty} (\sqrt[n]{2} - 1)^n \) was examined through the lens of convergence tests, revealing that such series can sometimes converge, even if their individual terms seem insignificant.

Approximation Methods

Approximation methods are useful techniques in mathematics to simplify complex expressions into more manageable forms. These methods allow us to assume that for large \( n \), terms simplify according to certain patterns or rules.

In this exercise, we approximated \( \sqrt[n]{2} \) as \( 1 + \frac{\ln(2)}{n} \) for large \( n \). This simplification made it possible to use convergence tests effectively and understand the behavior of the series better. Approximations become incredibly valuable when exact values are difficult to handle or unnecessary for determining convergence or divergence.