Problem 50 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the sum of the series $$ \sum_{k=1}^{\infty} \frac{2^{k}}{\left(2^{k+1}-1\right)\left(2^{k}-1\right)} $$

Step-by-Step Solution

Answer

The sum of the series is \(-\frac{1}{2}\).

Step 1: Look at the Structure of the Series

Analyze the general term of the series \( \frac{2^k}{(2^{k+1}-1)(2^k-1)} \). Notice that it appears to be in a telescoping form because the denominator can be decomposed into two factors that resemble partial fraction decomposition.

Step 2: Partial Fraction Decomposition

Perform a partial fraction decomposition on the term: \[ \frac{2^k}{(2^{k+1}-1)(2^k-1)} = \frac{A}{2^{k+1}-1} + \frac{B}{2^k-1} \] Multiply through by the denominator \((2^{k+1}-1)(2^k-1)\) to get an equation in terms of \(A\) and \(B\) after cross-multiplication:\[ 2^k = A(2^k-1) + B(2^{k+1}-1) \].

Step 3: Solve for A and B

Match coefficients to solve for \(A\) and \(B\). 1. Expand the equation: \[ A(2^k - 1) + B(2^{k+1} - 1) = A \cdot 2^k - A + B \cdot 2^{k+1} - B \] 2. Solve for \(B\) using \(B \cdot 2^{k+1} = 2^k\): \[ B = \frac{1}{2} \] 3. Substitute \(B\) back to find \(A\): \[ A = -\frac{1}{2} \]

Step 4: Write the Series using Found A and B

Using \( A = -\frac{1}{2} \) and \( B = \frac{1}{2} \), write the series term:\[ \frac{2^k}{(2^{k+1}-1)(2^k-1)} = \frac{-1}{2(2^k-1)} + \frac{1}{2(2^{k+1}-1)} \].This will help reveal the telescoping nature of the series.

Step 5: Identify the Telescoping Pattern

Observe the terms:\[ \sum_{k=1}^{\infty} \left( \frac{-1}{2(2^k-1)} + \frac{1}{2(2^{k+1}-1)} \right) \].When expanded, the series telescopes, meaning most intermediate terms will cancel each other out.

Step 6: Calculate the Sum

Telescoping results in cancellation:\[ -\frac{1}{2(2^1-1)} + \frac{1}{2(2^2-1)} - \frac{1}{2(2^2-1)} + \ldots \]Therefore, the series simplifies to the initial terms that do not have pairs:\[ -\frac{1}{2} \].Hence, the sum of the infinite series is \(-\frac{1}{2}\).

Key Concepts

Telescoping Series

In the world of infinite series, a telescoping series is a special type where most terms cancel each other out. This happens when consecutive terms in a series are structured such that additions and subtractions leave out only a few non-cancelled terms. The series we examined:

• Involves subtracting and adding fractions.
• The partial fractions hint at a telescoping nature.

When expanded, you'll see terms in the series cancel each other in pairs, making it simpler to find the sum of an otherwise complex series. By focusing on initial and final terms that do not cancel, you can efficiently determine the sum.

Partial Fraction Decomposition

Partial fraction decomposition is a tool used to break down complex fractions into simpler parts, often applied in calculus to evaluate series or integrals. In the exercise, we used this method to express:

• \( \frac{2^k}{(2^{k+1}-1)(2^k-1)} \)
• As a sum of two fractions: \( \frac{A}{2^{k+1}-1} + \frac{B}{2^k-1} \)

This allowed us to separate the original complex term into simpler fractions. By matching terms, we solved for unknowns \(A\) and \(B\), enabling us to apply the telescoping pattern effectively. This step is crucial in simplifying the series for further analysis.

Series Sum

Finding the sum of an infinite series can be daunting, but telescoping creates an opportunity to simplify this process. Here's how it works:

• Identify terms that cancel each other in the expansion.
• Focus on the first and last non-cancelled terms.

This method reduces the series to a sum of remaining terms, drastically simplifying calculations. In this specific example, it boils down to calculating the sum of a small number of terms. The net result was the sum of the series being \( -\frac{1}{2} \). This approach is powerful for series where recognition of patterns leads to quick solutions.

Mathematical Induction

Mathematical induction is a technique often used to prove mathematical statements for all natural numbers. Although not explicitly used in this solution, understanding induction can help foster comprehension of series properties. It consists of:

• Establishing a base case.
• Proving that if it holds for one case, it holds for the next.

In analyzing series, induction can be applied to confirm that specific properties or patterns hold true for all terms, ensuring validity of observations about behavior across infinite terms. It's crucial in rigorously demonstrating that the method of simplification, such as telescoping, consistently leads to a correct sum in an infinite series.