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Magazine Chapter 9: Problem 17
Finding the Interval of Convergence In Exercises \(15-38\) , find the interval of convergence of the power series. (Be sure to include a check for convergence at the endpoints of the interval.) $$\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{n}}{n}$$
Short Answer
Expert verified
The interval of convergence of the given power series is \(-1 < x \leq 1\).
Step by step solution
01
Setup the Ratio Test
The Ratio Test states that the power series \[\sum_{n=1}^{\infty} a_n\] converges if the limit \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L < 1\]. Apply this test to our series. The general term is \(a_n = \frac{(-1)^{n} x^{n}}{n}\), and \(a_{n+1} = \frac{(-1)^{n+1} x^{n+1}}{n+1}\).
02
Compute the Ratio
Next, compute the ratio \(\left| \frac{a_{n+1}}{a_n} \right|\). First, replace \(a_n\) and \(a_{n+1}\) with their corresponding expressions. Then, simplify the ratio. Doing so, we get \[\left| \frac{(-1)^{n+1} x^{n+1}/(n+1)}{(-1)^{n} x^{n}/n} \right| = \left| \frac{x(n)}{(n+1)} \right|\]. Remove the absolute value and our ratio then becomes \( \frac{x|n|}{n+1} \).
03
Take Limit
Take the limit of the ratio as \(n \to \infty\), to find the value of \(L\). We get \[\lim_{n \to \infty} \frac{x|n|}{n+1} = |x|\]. The series converges if \(L = |x| < 1\) and diverges if \(L = |x| > 1\).
04
Determine the Interval of Convergence
From the Ratio Test, the series converges when \(|x| < 1\), which yields the interval \(-1 < x < 1\). We must still confirm the endpoints for convergence. When \(x=1\), the series becomes \[\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n}\] which is an alternating series (also convergent by Leibniz鈥檚 theorem). Similarly, when \(x=-1\), the series becomes \[\sum_{n=1}^{\infty} \frac{(-1)^{2n}}{n} = \sum_{n=1}^{\infty} \frac{1}{n}\], which is a harmonic series (is a divergent series). Hence, it does not converge. So, the series converges only when \(x=1\) and so the interval is \(-1 < x \leq 1\).
05
Conclusion
Therefore, the interval of convergence of the given power series is \(-1 < x \leq 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series
A power series is a series of the form \[ \sum_{n=0}^{\infty} a_n (x-c)^n \]where \( a_n \) represents the coefficient for each term, \( x \) is the variable, and \( c \) is the center of the series. In this case, our series is \[ \sum_{n=1}^{\infty} \frac{(-1)^{n} x^{n}}{n} \] which does not have a \( c \) term, meaning it is centered at zero.
The power series is a fundamental concept in calculus and analysis. It allows functions to be expressed and analyzed in terms of an infinite sum of terms with powers of \( x \). The main focus when working with power series is to determine their convergence, especially understanding the interval over which the series converges.
The power series is a fundamental concept in calculus and analysis. It allows functions to be expressed and analyzed in terms of an infinite sum of terms with powers of \( x \). The main focus when working with power series is to determine their convergence, especially understanding the interval over which the series converges.
- Each term in the power series depends on the power \( n \), which increases indefinitely.
- The coefficients \( a_n \) and their behavior govern convergence properties.
Ratio Test
The Ratio Test is a key tool for determining the interval of convergence of a power series. The test tells us that a series \[ \sum_{n=1}^{\infty} a_n \] converges absolutely if \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L < 1 \].
For our series, we apply the Ratio Test by calculating the limit
In our solution, the limit \( |x| \) determines the convergence, showing that the series converges when \( |x| < 1 \).
For our series, we apply the Ratio Test by calculating the limit
- Determine \( a_{n+1} \) and \( a_n \) terms.
- Form the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \).
- Simplify and take the limit as \( n \to \infty \).
In our solution, the limit \( |x| \) determines the convergence, showing that the series converges when \( |x| < 1 \).
Convergence
Convergence refers to whether a series approaches a finite limit as more terms of the series are added. In the context of power series, a series can converge absolutely, conditionally, or diverge.
In our series example, the convergence behavior is determined primarily using the Ratio Test, showing absolute convergence between the interval \(-1 < x < 1\), with special consideration for the endpoints.
- Absolute Convergence: Occurs when the series \( \sum |a_n| \) converges. It guarantees overall convergence.
- Conditional Convergence: When a series converges but \( \sum |a_n| \) does not. An alternating series could converge conditionally.
- Divergence: Happens when the series does not reach a finite limit.
In our series example, the convergence behavior is determined primarily using the Ratio Test, showing absolute convergence between the interval \(-1 < x < 1\), with special consideration for the endpoints.
Alternating Series
The given series \[ \sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n} \] is an alternating series because the terms alternate in sign, thanks to the \((-1)^n\) factor. This type of series can converge even if its non-alternating counterpart does not.
We use the Alternating Series Test to check convergence at specific points, such as \( x = 1 \). The test states that an alternating series converges if:
We use the Alternating Series Test to check convergence at specific points, such as \( x = 1 \). The test states that an alternating series converges if:
- The absolute value of the terms \( |a_n| \) decreases monotonically.
- The terms approach zero, \( \lim_{n \to \infty} a_n = 0 \).
Harmonic Series
The harmonic series is an important concept in determining convergence at particular endpoints within an interval. The harmonic series is expressed as:\[ \sum_{n=1}^{\infty} \frac{1}{n} \]and is known for its divergence.
In our exercise, when evaluating the power series at \( x = -1 \), it becomes \[ \sum_{n=1}^{\infty} \frac{1}{n} \].Since the harmonic series does not converge, this implies that at \( x = -1 \), our original power series also diverges.
In our exercise, when evaluating the power series at \( x = -1 \), it becomes \[ \sum_{n=1}^{\infty} \frac{1}{n} \].Since the harmonic series does not converge, this implies that at \( x = -1 \), our original power series also diverges.
- A diverging harmonic series directly impacts endpoint convergence results.
- Knowledge of this divergence helps ensure the precision of establishing convergence intervals.
