Problem 27 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The series diverges.

Step 1: Test for Convergence using the Limit Comparison Test

Consider the series \( rac{k \ln k}{(k+1)^3} \). To apply the limit comparison test, we choose a comparison series \( b_k = \frac{k}{k^3} = \frac{1}{k^2} \). This is a \( p \)-series with \( p = 2 \), which is known to converge because \( p > 1 \).

Step 2: Compute the Limit for Comparison

Compute the limit \( \lim_{k \to \infty} \frac{a_k}{b_k} \) where \( a_k = \frac{k \ln k}{(k+1)^3} \) and \( b_k = \frac{1}{k^2} \). This gives:\[ \lim_{k \to \infty} \frac{ \frac{k \ln k}{(k+1)^3} }{ \frac{1}{k^2} } = \lim_{k \to \infty} \frac{k^3 \ln k}{(k+1)^3} \]Simplifying, we consider:\[ \lim_{k \to \infty} \left( \frac{k}{k+1} \right)^3 \ln k \]

Step 3: Simplify the Limit Expression

Calculate further:\[ \lim_{k \to \infty} \left( \frac{k}{k+1} \right)^3 \ln k = \lim_{k \to \infty} \left( 1 - \frac{1}{k+1} \right)^3 \ln k \]Since \( \left( 1 - \frac{1}{k+1} \right)^3 \to 1 \) as \( k \to \infty \), and \( \ln k \to \infty \), this limit simplifies to infinity. Therefore, the limit diverges.

Step 4: Conclusion on Divergence

Since the limit \( \lim_{k \to \infty} \frac{a_k}{b_k} \) is infinity, the limit comparison test indicates that the original series \( \sum_{k=1}^{\infty} \frac{k \ln k}{(k+1)^3} \) diverges.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a handy tool to determine if a series converges or diverges. It simplifies the process by contrasting an unknown series with a series whose behavior is known. Here's how it works:

• We have two series, say \( a_k \) and \( b_k \).
• The key idea is to compute \( \lim_{k \to \infty} \frac{a_k}{b_k} \).
• If the limit is a finite, non-zero number, then both series either converge or diverge together.

To effectively use the Limit Comparison Test, selecting the right comparison series, \( b_k \), is crucial. In our problem, \( b_k = \frac{1}{k^2} \) is chosen because it resembles our original series' form and it's a well-understood \( p \)-series. When using this method, remember that finding the right \( b_k \) is almost like a magic trick. It turns a seemingly complicated task into a straightforward calculation, making your work easier and more direct.

p-series

A \( p \)-series is a specific type of series that forms as \( \sum \frac{1}{k^p} \). It's one of the cornerstones of series and convergence tests. The behavior of a \( p \)-series depends heavily on the value of \( p \):

• If \( p > 1 \), then the series converges.
• If \( p \leq 1 \), the series diverges.

In our exercise, the comparison series \( \frac{1}{k^2} \) is a \( p \)-series with \( p = 2 \). Knowing \( p=2 > 1 \), we conclude this \( p \)-series converges. The \( p \)-series is a classic and straightforward way to know more about the nature of a series. It acts like a benchmark, helping us quickly decide the behavior of similar forms or those that seem slightly tricky to deduce on their own.

Convergence

Convergence is when the sum of an infinite series hits a specific value rather than going on indefinitely. It's similar to an infinitely long list of numbers' total being a number you can point at. In contrast, for divergence, you're chasing a sum that'll never land anywhere concrete. Series convergence ensures that as you sum more terms, they approach a precise value instead of climbing endlessly.

• In mathematical terms, if \( \lim_{n \to \infty} S_n = L \) and \( L \) is a real number, the series converges.
• A convergent series doesn't have to be evaluated against infinity; it settles at a finite result.

Understanding convergence helps interpret whether the series' behavior reaches an endpoint or dwindles without end. The original step in our exercise illustrates that by closely analyzing the \( p \)-series, the convergence test aligns with the series under test.

Divergence

Divergence occurs when the sum of a series continues to grow indefinitely without settling on a particular number. This signifies a series won't conclude neatly. The limit of the series' sum (if it existed) would carry on towards infinity, or it's oscillating without getting anywhere.

• A series diverges if \( \lim_{n \to \infty} S_n \) doesn't exist or is infinite.
• In layman's terms, think of a diverging series as running away without ever stopping.

In the exercise provided, the divergence is confirmed via the Limit Comparison Test. When the limit comparison led to infinity, we understood our original series would also diverge. Understanding the path of divergence tells us the series isn't bounded—it doesn't make a neat or finite jump to a sum.