Problem 24 - Calculus: Early Transcendentals

Question

Determine whether the series is convergent or divergent. \( \displaystyle \sum_{k = 1}^{\infty} ke^{-k^2} \)

Step-by-Step Solution

Answer

The series \( \sum_{k = 1}^{\infty} ke^{-k^2} \) is convergent.

Step 1: Understanding the Series

We are given the series \( \sum_{k = 1}^{\infty} ke^{-k^2} \). This is an infinite series where each term is of the form \( a_k = ke^{-k^2} \). We need to determine if this series converges or diverges.

Step 2: Setting up the Comparison Test

The given series includes the term \( e^{-k^2} \), an exponentially decreasing function, and \( k \), a linearly increasing function. To determine convergence, we can consider a comparison test with another series that might have similar behavior.

Step 3: Choose a Comparison Series

We can compare the given series with \( \sum_{k = 1}^{\infty} e^{-k^2} \). If \( \sum e^{-k^2} \) converges, then \( \sum ke^{-k^2} \) could also converge since \( k \) grows linearly while \( e^{-k^2} \) decreases exponentially faster.

Step 4: Test and Determine Convergence of Comparison Series

The series \( \sum_{k = 1}^{\infty} e^{-k^2} \) can be shown to converge by noting how fast \( e^{-k^2} \to 0 \) as \( k \to \infty \). The decrease is faster than any polynomial growth, leading to its convergence.

Step 5: Conclusion using the Comparison Test

Since \( \sum e^{-k^2} \) converges and \( ke^{-k^2} \leq C e^{-k^2} \) for a constant \( C \) when \( k \) is large enough, by the Comparison Test, \( \sum ke^{-k^2} \) also converges.

Key Concepts

Comparison Test

The Comparison Test is a fundamental tool in analyzing the convergence or divergence of infinite series. This test operates by comparing a given series to a second, "comparison" series, which is easier to evaluate.
By examining the behavior of the simpler series, we make deductions about the original series in question.
To apply the Comparison Test, find a series \(\sum_{k=1}^{\infty} b_k\) such that:

• The terms are non-negative: \(0 \leq a_k \leq b_k\) for all \(k\).
• If \(\sum b_k\) converges, then \(\sum a_k\) must also converge.
• If \(\sum b_k\) diverges, then \(\sum a_k\) must also diverge.

In our exercise, \(\sum_{k=1}^{\infty} ke^{-k^2}\), we chose \(\sum_{k=1}^{\infty} e^{-k^2}\) as the comparison series. This is because the exponential function decreases rapidly, providing a strong candidate for comparison.

Exponentially Decreasing Function

An exponentially decreasing function describes a process where the value of the function decreases at a rate proportional to its current value. It is typically represented as \(e^{-x}\).
When interpreting this behavior within the context of series, \(e^{-k^2}\) in the expression \(ke^{-k^2}\) decreases more quickly than \(k\) grows since the square of any number becomes much larger faster than its linear counterpart.
This rapid decline in \(e^{-k^2}\)'s value is significant in understanding convergence, as such functions tend towards zero more quickly than polynomial functions, providing a strong basis for proving series convergence.

Infinite Series

An infinite series is a sum of an infinite sequence of terms. Specifically, it involves summing terms \(a_k\) from \(k=1\) to infinity. The challenge with infinite series lies in determining whether their sums converge (approach a specific value) or diverge (grow without bound).
Convergence means that as you add more terms, the total approaches a specific number, whereas divergence means the series goes to infinity or does not settle at one value.
The primary focus is often on understanding how quickly the individual terms \(a_k\) tend to zero as k increases, since this is crucial for determining convergence.

Polynomial Growth

Polynomial growth refers to expressions of the form \(k^n\) where \(n\) is a positive integer. They grow at a slower rate compared to exponential functions like \(e^{-k^2}\). This growth can be visualized as creating terms that increase steadily with each increment of \(k\).
In the given series \(\sum_{k=1}^{\infty} ke^{-k^2}\), the \(k\) term represents linear polynomial growth. Comparing this polynomial behavior to the exponential decay term \(e^{-k^2}\) provides insight into the series' potential convergence.
The balance between polynomial growth and exponential decay plays an integral role in predicting the behavior of series as \(k\) becomes large.