Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { Show that } \sum_{k=2} \ln \left(1-\frac{1}{k^{2}}\right)=-\ln 2 $$

Step-by-Step Solution

Answer

The series sums to \(-\ln 2\).

Step 1: Express Single Term with Logarithm Property

First, let's use the property of logarithms: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). We can rewrite each term \( \ln\left(1-\frac{1}{k^2}\right) \) as \( \ln\left(\frac{k^2-1}{k^2}\right) \). This simplifies to \( \ln\left(\frac{(k-1)(k+1)}{k^2}\right) \).

Step 2: Expand the Series Using Logarithmic Identity

Now substitute the expression from Step 1 into the series. We need to evaluate the series:\[\sum_{k=2} \ln\left(\frac{(k-1)(k+1)}{k^2}\right).\]Using the properties of logarithms, this becomes:\[\sum_{k=2} \left(\ln(k-1) + \ln(k+1) - 2\ln(k)\right).\]

Step 3: Examine the Telescoping Nature of the Series

Notice the series is telescoping. Write out the first few terms:\[\ln(1) + \ln(3) - 2\ln(2) + \ln(2) + \ln(4) - 2\ln(3) + \ln(3) + \ln(5) - 2\ln(4) + ...\]This results in many terms canceling with each other. Specifically, each \( \ln(k+1) \) term is canceled by the negative \( 2\ln(k) \) and then add \( \ln(k-1) \) in the next step.

Step 4: Simplify the Remaining Terms

Upon canceling, the only terms that remain uncanceled are the initial\( \ln(1) \) and the remaining term from the last segment of the series \( - \ln(n+1) \) as \( n \to \infty \). However, practically, only terms around the first values are not canceled. Thus:\[\sum_{k=2}\ln\left(1-\frac{1}{k^2}\right) = \ln(1) - \ln(\infty) \, + \, \ln(2)\] As \( \ln(1) = 0 \), cancelation leads to the remaining uncanceled term being summed to \(-\ln(2)\).

Step 5: Conclude the Summation Result

Our final remaining term, after the telescopic cancellation, is \(-\ln(2)\). Thus, the entire summation reduces to:\[\sum_{k=2} \ln\left(1-\frac{1}{k^2}\right) = -\ln 2.\]This completes the verification of the given series.

Key Concepts

Properties of Logarithms

Logarithms are mathematical operations that are the inverse of exponentiation, and they have several important properties that make them useful in algebra and calculus. These properties allow us to transform complex expressions into simpler forms. Here are some key logarithmic properties that are essential to understand:

• Product Property: \( \ln(a \cdot b) = \ln(a) + \ln(b) \)
• Quotient Property: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
• Power Property: \( \ln(a^b) = b \cdot \ln(a) \)

This exercise utilizes the quotient property to simplify each term within the series. By expressing terms in the form \( \ln\left(\frac{(k-1)(k+1)}{k^2}\right) \), we can further decompose the series into a more manageable summation, highlighting the use of these properties in series manipulation.

Series Summation

Series summation involves adding up the terms of a sequence, a fundamental concept in calculus and analysis. In this context, each individual term in the series is transformed using logarithmic properties to create a telescoping series. A telescoping series is one where most terms cancel each other out, simplifying the computation greatly.
For the given series:

• Transform each term using logarithm properties.
• Recognize patterns that lead to cancellation.
• Simplify the remaining expression.

The original series \[ \sum_{k=2} \ln\left(1-\frac{1}{k^2}\right) \] utilizes the properties of logarithms to convert into a telescoping nature, dramatically simplifying the entire summation process by canceling intermediate terms.

Convergence of Infinite Series

In mathematics, convergence refers to the idea that a series approaches a specific value as the number of terms goes to infinity. For our series:\[ \sum_{k=2} \ln\left(1-\frac{1}{k^2}\right) \]we observe convergence by analyzing how the terms behave as \( k \to \infty \).
The telescoping nature ensures that most terms cancel out, leaving a finite remaining value, typically involving the first and last terms. Here are steps to understand convergence:

• Identify the behavior of each term as \( k \) increases.
• Analyze any remaining terms after extensive cancellation.
• Verify the limit of the uncanceled terms, confirming convergence to a known value.

This series converges to \( -\ln(2) \), exhibiting how a telescoping series can neatly settle into a finite value despite initially infinite terms.

Mathematical Proof Techniques

Mathematical proofs are foundational to validate propositions and findings within mathematics. The exercise employs techniques like algebraic manipulation and analysis of series properties to derive a result from an infinite series.
Understanding proof techniques involves:

• Breaking down complex expressions into understandable parts using known identities, like logarithm properties.
• Applying logical reasoning to simplify and solve problems.
• Utilizing properties such as telescoping to identify and eliminate terms systematically.

The step-by-step approach showcased in our solution strengthens the understanding of how to handle and prove elements of a series. This comprehensive method ensures students grasp the logical progression from an infinite series to its convergent outcome.