Problem 10 - Calculus One Again

Question

Let \(\boldsymbol{x} \in \mathbb{R} \backslash\\{\mathbf{1}\\}\) and let \(\boldsymbol{n} \in \mathbb{N} \backslash\\{\mathbf{0}\\}\). Prove that \(\sum_{k=0}^{n} \frac{2^{k}}{x^{2^{k}}+1}=\frac{1}{x-1}-\frac{2^{n+1}}{x^{2^{n+1}}+1}\)

Step-by-Step Solution

Answer

The given equation is true by telescoping series simplification.

Step 1: Understand the Problem

We need to prove the equation \( \sum_{k=0}^{n} \frac{2^{k}}{x^{2^{k}}+1} = \frac{1}{x-1}-\frac{2^{n+1}}{x^{2^{n+1}}+1} \). The key here is to express this sum as something telescopic, which simplifies when summed.

Step 2: Rearrange the Summation

Rewrite each term \( \frac{2^{k}}{x^{2^{k}}+1} \) as \( \frac{1}{x-1} \left( \frac{1}{x^{2^{k}}} - \frac{1}{x^{2^{k+1}}} \right) \). This form is obtained by algebraic manipulation, with the goal of setting up a telescoping series.

Step 3: Recognize the Telescoping Series

When terms are arranged as \( \frac{1}{x-1} \left( \frac{1}{x^{2^{k}}} - \frac{1}{x^{2^{k+1}}} \right) \), notice how successive terms cancel: only the first and the last terms remain when the summation is expanded.

Step 4: Simplify the Summation

Calculate the sum: The first term is \( \frac{1}{x^{0} \cdot (x-1)} = \frac{1}{x-1} \) and the last one, coming after all intermediate terms cancel, is \( \frac{-1}{x^{2^{n+1}} \cdot (x-1)} \).

Step 5: Final Expression

Combine the result from the telescoping series: \( \sum_{k=0}^{n} \left( \frac{1}{x^{2^{k}}} - \frac{1}{x^{2^{k+1}}} \right) = \frac{1}{x-1} - \frac{1}{x^{2^{n+1}}(x-1)} \), and factor the last term with the powers of 2. Hence, express the result as \( \frac{1}{x-1} - \frac{2^{n+1}}{x^{2^{n+1}}+1} \).

Key Concepts

Algebraic Manipulation

In mathematics, algebraic manipulation is a technique used to rearrange expressions in order to solve equations or simplify complex expressions. It involves the use of algebraic properties such as distributive, associative, and commutative properties, alongside techniques like factoring, expanding, and canceling.

When dealing with the given exercise, algebraic manipulation becomes vital. We start by considering how each term of the series can be rearranged. By cleverly rewriting the given expression, \( \frac{2^{k}}{x^{2^{k}}+1} \), our goal is to transform it into a format that will allow the series to telescope.

This involves expressing each term as \( \frac{1}{x-1} \left( \frac{1}{x^{2^k}} - \frac{1}{x^{2^{k+1}}} \right) \). This transformation simplifies the complex fraction into a difference of fractions, prepared to unfold into a telescoping series when summed. The manipulation not only makes the calculations more manageable, but it also reveals the elegant structure lying beneath the problem.

Summation Techniques

Summation techniques are crucial tools in calculus and analysis for finding the total of terms in a series. This exercise showcases a specific technique known as telescoping series.

A telescoping series is one in which most terms cancel out when the whole series is summed, leaving only a few terms with significant values. In the sum \( \sum_{k=0}^{n} \frac{2^{k}}{x^{2^{k}}+1} \), each term is rewritten to reveal a cancellation pattern: \( \frac{1}{x-1} \left( \frac{1}{x^{2^{k}}} - \frac{1}{x^{2^{k+1}}} \right) \).

When summed, the terms inside the parentheses systematically cancel each other out, except for the very first term of the first bracket and the very last term of the last bracket.

• The first surviving term: \( \frac{1}{x-1} \)
• The last surviving term: \( -\frac{1}{x^{2^{n+1}}(x-1)} \)

This cancellation simplifies our work drastically, allowing us to acquire a simple closed-form expression for the series.

Mathematical Proofs

Mathematical proofs are logical arguments that demonstrate the truth of a mathematical statement beyond doubt. Every mathematical proof relies heavily on a logical sequence and theorems, and in our exercise, the proof is achieved through recognizing and exploiting a telescoping pattern.

First, we establish the initial form and transform it using algebraic manipulation to reveal its telescopic nature. The series reduction, only showing the non-cancelled start and end terms, becomes evident through careful analysis.

Next, we show that by summing our transformed series, \( \frac{1}{x-1} - \frac{2^{n+1}}{x^{2^{n+1}}+1} \), we achieve an equivalent statement to the provided sum. This step-by-step reduction articulated clearly, offers a complete and concise proof of the given equation.

Proving mathematical statements, like this one, underscores the power of structured thinking, logical progression, and the effective use of established mathematical techniques such as algebraic manipulation and series summation.