Chapter 9: Problem 9
Find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. \(\frac{x}{1 \cdot 2}-\frac{x^{2}}{2 \cdot 3}+\frac{x^{3}}{3 \cdot 4}-\frac{x^{4}}{4 \cdot 5}+\frac{x^{5}}{5 \cdot 6}-\cdots\)
Short Answer
Expert verified
The convergence set is \([-1, 1)\).
Step by step solution
01
Identify and Express the General Term
The given series is \( \frac{x}{1 \cdot 2} - \frac{x^2}{2 \cdot 3} + \frac{x^3}{3 \cdot 4} - \frac{x^4}{4 \cdot 5} + \cdots \). Notice that each term of the series is of the form \( (-1)^{n+1} \cdot \frac{x^n}{n(n+1)} \). Thus, we can express the general term as \( a_n = (-1)^{n+1} \cdot \frac{x^n}{n(n+1)} \).
02
Apply the Absolute Ratio Test
The Absolute Ratio Test states that a series \( \sum a_n \) converges if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). Substitute the general term and calculate:\[\frac{a_{n+1}}{a_n} = \frac{(-1)^{n+2} \cdot \frac{x^{n+1}}{(n+1)(n+2)}}{(-1)^{n+1} \cdot \frac{x^n}{n(n+1)}} = \frac{x \cdot n}{(n+2)}\]Thus, \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x \cdot n}{(n+2)} \right| \).
03
Find the Limit
Now, take the limit as \( n \to \infty \):\[\lim_{n \to \infty} \left| \frac{x \cdot n}{n+2} \right| = \lim_{n \to \infty} \frac{|x| \cdot n}{n+2}\]Simplifying gives:\[\lim_{n \to \infty} \frac{|x| \cdot n}{n+2} = \lim_{n \to \infty} \frac{|x| \cdot n}{n} = |x|\].
04
Establish the Convergence Criterion
For the series to converge, the result from the Ratio Test must be less than 1:\[ |x| < 1 \]This implies that the series converges for \( -1 < x < 1 \).
05
Verify the Endpoints
To determine the convergence set, check the behavior at the endpoints \( x = -1 \) and \( x = 1 \). - **For \( x = 1 \)**: The series becomes \( \sum (-1)^{n+1} \cdot \frac{1}{n(n+1)} \), which converges by the Alternating Series Test. - **For \( x = -1 \)**: The series becomes \( \sum \frac{(-1)^n}{n(n+1)} \), also convergent by the Alternating Series Test.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Absolute Ratio Test
The Absolute Ratio Test is a powerful method for determining the convergence of a series. It is often used when a series consists of complex terms and involves non-obvious patterns.
In our original problem, we used \( a_n = (-1)^{n+1} \cdot \frac{x^n}{n(n+1)} \) and found the ratio to simplify to \( \left| \frac{x \cdot n}{n+2} \right| \), leading to the condition \( |x| < 1 \) for convergence.
- This test examines the limit of the absolute value of the ratio between consecutive terms in a series.
- The series converges absolutely when this limit is less than 1.
In our original problem, we used \( a_n = (-1)^{n+1} \cdot \frac{x^n}{n(n+1)} \) and found the ratio to simplify to \( \left| \frac{x \cdot n}{n+2} \right| \), leading to the condition \( |x| < 1 \) for convergence.
Alternating Series Test
The Alternating Series Test provides a criterion for convergence for series where the terms alternate in sign.
This test is very useful for series with terms like \((-1)^{n} \frac{1}{n+1}\). Due to the alternation of signs, the series converges even if the absolute values of the terms do not.
In our exercise, for both \( x = 1 \) and \( x = -1 \), the series terms decrease and approach zero, which ensures convergence of the series at these endpoints using the Alternating Series Test.
- For a series \( \sum (-1)^n b_n \), the test can be applied if two conditions are met:
- The sequence \( b_n \) is decreasing.
- \( \lim_{n \to \infty} b_n = 0 \).
This test is very useful for series with terms like \((-1)^{n} \frac{1}{n+1}\). Due to the alternation of signs, the series converges even if the absolute values of the terms do not.
In our exercise, for both \( x = 1 \) and \( x = -1 \), the series terms decrease and approach zero, which ensures convergence of the series at these endpoints using the Alternating Series Test.
General Term of a Series
The general term of a series, denoted \( a_n \), is crucial for understanding the nature of the series. It represents the formula that generates each term of the series from natural numbers \( n \).
- The general term helps in checking for convergence using various tests.
- Knowing \( a_n \) allows the reconstruction of the entire series for any value of \( n \).
- This is derived from the pattern or sequence present in the given series.
Limits in Calculus
Limits are a fundamental concept in calculus used to understand the behavior of functions as inputs approach a certain value. In the context of series convergence:
In our series, computing the limit \( \lim_{n \to \infty} \frac{|x| \cdot n}{n+2} \) resulted in simplifying the problem to \( |x| \). This limit indicated the range of \( x \) values for which the series converges. Understanding this principle of limits allows us to identify whether a series diverges or converges based on its input value.
- Limits help determine the long-term behavior of sequences and series.
- They are vital for applying tests like the Absolute Ratio Test and others.
In our series, computing the limit \( \lim_{n \to \infty} \frac{|x| \cdot n}{n+2} \) resulted in simplifying the problem to \( |x| \). This limit indicated the range of \( x \) values for which the series converges. Understanding this principle of limits allows us to identify whether a series diverges or converges based on its input value.