Problem 15 - Calculus: Early Transcendentals

Question

Test the series for convergence or divergence. \( \displaystyle \sum_{k = 1}^{\infty} \frac {2^{k-1} 3^{k+1}}{k^k} \)

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Simplify the Expression

First, simplify the general term of the series: \( a_k = \frac {2^{k-1} 3^{k+1}}{k^k} \). Rewrite this expression as \( a_k = \frac{2^{k-1} \cdot 3^{k+1}}{k^k} = \frac{2^{k-1} \cdot 3 \cdot 3^k}{k^k} = 3 \cdot \frac{2^{k-1} \cdot 3^k}{k^k} \).

Step 2: Apply the Ratio Test

Use the Ratio Test to determine convergence or divergence. This requires analyzing the limit \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \). Substitute \( a_{k+1} = 3 \cdot \frac{2^k \cdot 3^{k+1}}{(k+1)^{k+1}} \) and \( a_k = 3 \cdot \frac{2^{k-1} \cdot 3^k}{k^k} \).

Step 3: Simplify the Ratio

Calculate \( \frac{a_{k+1}}{a_k} = \frac{3 \cdot \frac{2^k \cdot 3^{k+1}}{(k+1)^{k+1}}}{3 \cdot \frac{2^{k-1} \cdot 3^k}{k^k}} \). Simplify this as \( \frac{2 \cdot 2^k \cdot 3^k \cdot 3}{2^{k-1} \cdot 3^k} \cdot \frac{k^k}{(k+1)^{k+1}} \). Further simplifying gives: \( 6 \cdot \left(\frac{k}{k+1}\right)^k \cdot \frac{1}{k+1} \).

Step 4: Evaluate the Limit

Now, evaluate \( \lim_{k \to \infty} 6 \cdot \left( \frac{k}{k+1} \right)^k \cdot \frac{1}{k+1} \). Since \( \left( \frac{k}{k+1} \right)^k \to \frac{1}{e} \) and \( \frac{1}{k+1} \to 0 \) as \( k \to \infty \), the entire expression tends to 0.

Step 5: Conclude with the Ratio Test

The Ratio Test states that if \( \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \), then the series converges. Since the limit evaluates to 0, which is less than 1, the series converges.

Key Concepts

Ratio Test

When determining if a series converges or diverges, one of the most useful methods is the Ratio Test. This test is particularly handy when dealing with series involving factorials, exponentials, and powers. To apply this test, analyze the limit of the absolute value of the ratio of consecutive terms in the series:\[ \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \]Here's what you need to do:- Simplify the expression for \( a_{k+1} \) and \( a_k \).- Substitute these into the ratio \( \frac{a_{k+1}}{a_k} \).- Simplify the ratio as much as possible.After simplification, evaluate the limit of the reduced expression as \( k \to \infty \). If the resulting limit is less than 1, the series converges absolutely. If the limit is greater than 1, the series diverges. If it equals 1, the test is inconclusive.

Infinite Series

An infinite series is simply a sum of infinitely many terms. A familiar example is the series \( \sum_{k=1}^{\infty} a_k \), where \( a_k \) is a general term formula.But what does it mean for this series to converge? When we say an infinite series converges, it means that as you add more and more terms, the sum approaches a specific number, called the limit.- **Convergence**: The sum approaches a finite value.- **Divergence**: The sum does not approach any limit and either increases indefinitely or oscillates.Infinite series can arise in various mathematical contexts like calculus, sequences, and even in solving differential equations. Whether they converge or diverge often depends on the nature of the terms \( a_k \), hence why tests like the Ratio Test become invaluable tools.

Limit Evaluation

To solve many problems involving series, evaluating limits is crucial. Limit evaluation involves understanding what happens to terms of a sequence or function as the variable approaches some value or infinity.In our example, calculating the limit \( \lim_{k \to \infty} 6 \cdot \left( \frac{k}{k+1} \right)^k \cdot \frac{1}{k+1} \) is key to determining convergence using the Ratio Test.Here's the process broken down:- **Identify any patterns or simplifications**: Recognize any expressions that are well-known limits. For example, \( \left( \frac{k}{k+1} \right)^k \to \frac{1}{e} \) as \( k \to \infty \).- **Consider dominant terms in the limit**: Consider which term dominates the expression's behavior as \( k \to \infty \). In this case, \( \frac{1}{k+1} \to 0 \) simplifies the evaluation.Ultimately, understanding how to evaluate limits allows us to deeply understand the behavior of series and functions, determining if their infinite nature results in convergence to a specific value.