Chapter 9: Problem 11
Show that each series converges absolutely. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{1}{n(n+1)} $$
Short Answer
Expert verified
The series converges absolutely as \( \sum \frac{1}{n(n+1)} \) is convergent.
Step by step solution
01
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 \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n(n+1)} \).
02
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.
03
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.
04
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.
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
After establishing absolute convergence, we learn that each alternating series also demands the scrutiny of its general term properties.
- The terms \(a_n\) are positive.
- The terms are decreasing, i.e., \(a_{n+1} \leq a_n\) for all \(n\).
- \(\lim_{n \to \infty} a_n = 0\).
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:
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.
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:
For the fraction \(\frac{1}{n(n+1)}\), calculations result in:
- \(A = 1\)
- \(B = -1\)
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:
The convergence of series is a fundamental component of calculus, which underpins much of analysis in mathematics and its applications.
- Comparison Test
- Ratio Test
- Integral Test
- P-Series Test
The convergence of series is a fundamental component of calculus, which underpins much of analysis in mathematics and its applications.