Problem 19 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use any test developed so far, including any from Section \(9.2\), to decide about the convergence or divergence of the series. Give a reason for your conclusion. $$ \sum_{k=1}^{\infty} k^{2} e^{-k^{3}} $$

Step-by-Step Solution

Answer

The series converges by the Ratio Test because the limit is zero.

Step 1: Understand the Series

The given series is \( \sum_{k=1}^{\infty} k^{2} e^{-k^{3}} \). We need to determine whether this series converges or diverges by using any appropriate test.

Step 2: Identify the Convergence Criteria

Given the form of the terms \( a_k = k^{2} e^{-k^{3}} \), consider applying the Ratio Test as \( e^{-k^{3}} \) indicates possible rapid decay.

Step 3: Apply the Ratio Test

For the Ratio Test, compute \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \), where \( a_k = k^{2} e^{-k^{3}} \) and \( a_{k+1} = (k+1)^{2} e^{-(k+1)^{3}} \).

Step 4: Calculate the Limit in Ratio Test

Compute\[L = \lim_{k \to \infty} \left| \frac{(k+1)^{2} e^{-(k+1)^{3}}}{k^{2} e^{-k^{3}}} \right| = \lim_{k \to \infty} \left| \frac{(k+1)^{2}}{k^{2}} e^{-(k+1)^{3} + k^{3}} \right|\]Notice that the exponent simplifies to:\[-(k+1)^{3} + k^{3} = -(k^{3} + 3k^{2}+3k+1) + k^{3} = -3k^{2} - 3k - 1\]Thus,\[L = \lim_{k \to \infty} \left(1 + \frac{1}{k}\right)^2 \, e^{-3k^2 - 3k - 1}\]

Step 5: Simplify and Evaluate the Limit

As \( k \to \infty \), \( \left(1 + \frac{1}{k}\right)^2 \to 1 \) and \( e^{-3k^2 - 3k - 1}\) decreases exponentially towards zero. Therefore, \( L = 0 \).

Step 6: Conclusion from the Ratio Test

Since \( L < 1 \) (specifically \( L = 0 \)), the Ratio Test confirms that the series \( \sum_{k=1}^{\infty} k^{2} e^{-k^{3}} \) converges.

Key Concepts

Ratio Test

The Ratio Test is a popular method used to determine if a series converges or diverges. It is particularly useful when dealing with series where each term consists of factorials, exponential functions, or any form that exhibits rapid growth or decay.
This test compares the limit of the ratio of successive terms in a series as the term index approaches infinity. If you have a series \( \sum a_k \), the Ratio Test looks at the limit \( L = \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| \).

• If \( L < 1 \), the series converges.
• If \( L > 1 \) or is infinite, the series diverges.
• If \( L = 1 \), the test is inconclusive.

The test is handy because the exponential decay factor, such as \( e^{-k^3} \) in our original series, simplifies the evaluation of the limit significantly.

Convergent Series

A convergent series is one where the sum of its terms approaches a finite value as more terms are added. The idea is that as you add infinitely many terms, the total sum doesn't blow up to infinity. For a series \( \sum a_k \), convergence means that the partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) approach a finite number as \( n \to \infty \).
Some series have this property under certain tests, one being the Ratio Test. In our example, using \( L = 0 \), which is less than 1, confirmed that the series \( \sum_{k=1}^{\infty} k^{2} e^{-k^{3}} \) converges. A convergent series ensures that the operation of summation over infinitely many terms is stable, yielding a comprehensible result.

Exponential Function

The exponential function \( e^x \) is a significant mathematical concept, often appearing in series and differential equations. It represents growth or decay processes and is known for its unique property: the derivative of \( e^x \) is itself.
In our series, the exponential term \( e^{-k^3} \) acts as a decay factor. As \( k \) increases, \( e^{-k^3} \) decreases rapidly towards zero, which impacts the convergence of the series due to its fast-decaying rate. This decay is crucial because it counterbalances the growth of the other factor, \( k^2 \).
The nature of exponential functions is what allows rapidly increasing or decreasing terms in series to be managed effectively during convergence tests.

Series Convergence

Series convergence refers to the behavior of a series as the number of terms grows indefinitely. Identifying the convergence of a series is key in mathematical analysis, and various tests, like the Ratio Test, aid in this process.
For convergence, the infinite sum must not incrementally diverge to infinity. Instead, it should stabilize, reaching a fixed value. This requires a delicate balance between the terms of the series.
In the original exercise, we examined the series \( \sum_{k=1}^{\infty} k^{2} e^{-k^{3}} \). We found that rapid exponential decay outpaces polynomial growth, implying convergence. Understanding the underlying tests and behaviors is crucial in distinguishing between convergent and divergent series.