Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that each series converges absolutely. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n(n+1)} $$

Step-by-Step Solution

Answer

The series converges absolutely as \( \sum \frac{1}{n(n+1)} \) is convergent.

Step 1: Understand Absolute Convergence

A series \( \sum a_n \) converges absolutely if the series \( \sum |a_n| \) converges. In this exercise, we need to analyze the absolute value of the given alternating series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n(n+1)} \).

Step 2: Consider the Absolute Value of Terms

Consider the series of absolute values: \( \sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{1}{n(n+1)} \right| = \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \). We remove the alternating factor because absolute convergence depends on non-negative terms.

Step 3: Simplify the Terms

Notice that \( \frac{1}{n(n+1)} \) can be rewritten using partial fraction decomposition: \( \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \). This provides a way to simplify and analyze the series.

Step 4: Use Telescoping Series Properties

The series \( \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+1} \right) \) is a telescoping series. In such a series, many terms cancel out, leaving only a few terms from the beginning and end. Specifically, this series simplifies to \( 1 \), implying convergence.

Step 5: Conclude Absolute Convergence

Since the series \( \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \) converges (due to its telescoping nature), the original series \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n(n+1)} \) converges absolutely by the definition of absolute convergence.

Key Concepts

Alternating Series

An alternating series is a series where the signs of its terms alternate between positive and negative. In mathematical terms, it usually involves something of the form \[ (-1)^n a_n \]or \[ (-1)^{n+1} a_n. \]This change in sign can create an oscillating pattern. A typical example is the alternating harmonic series:\[\sum_{n=1}^\infty (-1)^{n+1} \frac{1}{n}. \]Alternating series can sometimes converge while the corresponding series of absolute values diverges. To determine convergence, one often uses the Alternating Series Test. For an alternating series \(\sum_{n=1}^\infty (-1)^n a_n\), the series converges if:

• The terms \(a_n\) are positive.
• The terms are decreasing, i.e., \(a_{n+1} \leq a_n\) for all \(n\).
• \(\lim_{n \to \infty} a_n = 0\).

Alternating series can lead to absolute or conditional convergence, making them an important function in series analysis.
After establishing absolute convergence, we learn that each alternating series also demands the scrutiny of its general term properties.

Telescoping Series

A telescoping series is a series where many terms cancel out between successive steps. This characteristic becomes very useful for finding the sum easily. An example of a telescoping series looks like this:\[\sum_{n=1}^\infty \left( a_n - a_{n+1} \right).\]This format allows internal terms to cancel, leaving only a sum of the initial and final terms after simplification.
In our exercise's context, partial fraction decomposition converts \[ \frac{1}{n(n+1)} \]into \[ \frac{1}{n} - \frac{1}{n+1}. \]Following this transformation, the sum of the series simplifies due to extensive cancellations:

• The initial element \(\frac{1}{1}\) remains.
• Subsequent terms cancel each other out.
• The final subtraction \(- \frac{1}{n+1}\) approaches zero as \(n\) approaches infinity.

The cancellation phenomenon makes the convergent result much clearer and calculable.

Partial Fraction Decomposition

Partial Fraction Decomposition is a method used particularly in algebra to break down complex rational expressions into simpler fractions. This approach simplifies integration and series resolution. For an expression like \(\frac{1}{n(n+1)}\), partial fraction decomposition involves writing:\[ \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}. \]Solving for \(A\) and \(B\) often involves equating coefficients from both sides of an equation and solving simultaneous equations.
For the fraction \(\frac{1}{n(n+1)}\), calculations result in:

• \(A = 1\)
• \(B = -1\)

Thus, the result \(\frac{1}{n} - \frac{1}{n+1}\) is achieved. Partial fraction decomposition makes telescopic cancellation possible, leading to easier computation of series.

Convergent Series

A series is said to be convergent if the sum of its infinite terms approaches a finite number. Convergence analyzes whether adding up all terms of the series leads to a sensible, finite result instead of growing infinitely large or oscillating erratically. For alternately or purely positive series, various tests can verify convergence, such as:

• Comparison Test
• Ratio Test
• Integral Test
• P-Series Test

In our exercise, the scenario illustrates absolute convergence, where \(\sum_{n=1}^{\infty} \frac{1}{n(n+1)}\) converges due to its simplified telescopic nature, leaving a manageable result after comprehensive cancellation.
The convergence of series is a fundamental component of calculus, which underpins much of analysis in mathematics and its applications.