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Magazine Chapter 9: Problem 50
Find the sum of the series $$ \sum_{k=1}^{\infty} \frac{2^{k}}{\left(2^{k+1}-1\right)\left(2^{k}-1\right)} $$
Short Answer
Expert verified
The sum of the series is \(-\frac{1}{2}\).
Step by step solution
01
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.
02
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:\[ 2^k = A(2^k-1) + B(2^{k+1}-1) \].
03
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 \(A\): \[ A = -\frac{1}{2} \]
04
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.
05
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.
06
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 \(-\frac{1}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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: \( \frac{A}{2^{k+1}-1} + \frac{B}{2^k-1} \)
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.
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.
