Problem 30
Question
Suppose that \( \sum_{n = 0}^{\infty} c_nx^n \) converges when \( x = - 4 \) and diverges when \( x = 6. \) What can be said about the convergence or divergence of the following series? (a) \( \sum_{n = 0}^{\infty} c_n \) (b) \( \sum_{n = 0}^{\infty} c_n8^n \) (c) \( \sum_{n = 0}^{\infty} c_n( - 3)^n \) (d) \( \sum_{n = 0}^{\infty} ( - 1)^n c_n 9^n \)
Step-by-Step Solution
Verified Answer
(a) Converges; (b) Diverges; (c) Converges; (d) Diverges.
1Step 1: Identify the Radius of Convergence
Since the series \( \sum_{n = 0}^{\infty} c_nx^n \) converges at \( x = -4 \) and diverges at \( x = 6 \), we can determine the radius of convergence \( R \) of the power series. The point \( x = -4 \) is inside the interval of convergence, while \( x = 6 \) is outside it. Therefore, the distance from the center of the interval (usually 0) to -4 gives a condition that \( |x| = 4 \) is within the interval. The divergence at \( x=6 \) means that \( |x|=6 \) is outside the interval, hence \( R < 6 \) and \( R \geq 4 \). This gives the radius of convergence as \( R = 5 \), as it must encompass 4 but not reach 6.
2Step 2: Consider the Series at Different Points
Using the radius of convergence \( R = 5 \), we analyze each given series to check if \( |x| \) falls within this radius or not, for convergence.(a) For \( \sum_{n = 0}^{\infty} c_n \), \( x = 1 \), so \( |x| = 1 < 5 \). Thus, the series converges.(b) For \( \sum_{n = 0}^{\infty} c_n8^n \), \( x = 8 \), so \( |x| = 8 > 5 \). Therefore, the series diverges.(c) For \( \sum_{n = 0}^{\infty} c_n(-3)^n \), \( x = -3 \), so \( |x| = 3 < 5 \). Hence, the series converges.(d) For \( \sum_{n = 0}^{\infty} (-1)^n c_n 9^n \), \( x = 9 \), so \( |x| = 9 > 5 \). Thus, the series diverges.
Key Concepts
Power SeriesConvergenceDivergenceInterval of Convergence
Power Series
A power series is a type of infinite series that is represented as a sum of terms made up of a coefficient and a variable raised to a power. It is usually expressed in the form:
Power series play a crucial role because they can be used to represent functions in a broad variety of fields, such as in calculus or complex analysis.
A key aspect of understanding power series is determining the range of \( x \) values where the series converges. This range is essential for ensuring the series accurately represents a function within a certain domain.
- \( \, \sum_{n=0}^{\infty} c_n(x-a)^n \)
Power series play a crucial role because they can be used to represent functions in a broad variety of fields, such as in calculus or complex analysis.
A key aspect of understanding power series is determining the range of \( x \) values where the series converges. This range is essential for ensuring the series accurately represents a function within a certain domain.
Convergence
Convergence in the context of power series happens when the sum of the terms approaches a specific finite value as the number of terms increases.
When we say a series converges, it implies that as you add more and more terms, the series approaches a limit, rather than continually increasing or decreasing.
For a power series \( \sum_{n=0}^{\infty} c_n(x-a)^n \) to converge at a specific \( x \), the absolute value of \( x-a \) must be less than the radius of convergence \( R \). Three useful tests to determine convergence include:
When we say a series converges, it implies that as you add more and more terms, the series approaches a limit, rather than continually increasing or decreasing.
For a power series \( \sum_{n=0}^{\infty} c_n(x-a)^n \) to converge at a specific \( x \), the absolute value of \( x-a \) must be less than the radius of convergence \( R \). Three useful tests to determine convergence include:
- Ratio Test: This test can help identify the range of \( x \) that makes the series converge.
- Root Test: Similar to the ratio test, it provides another approach to check series convergence.
- Direct Substitution: Plugging specific values of \( x \) into the series to see if the series converges.
Divergence
When a series does not converge, it diverges. This means that as more and more terms are added, the series does not approach a finite limit but instead increases indefinitely or oscillates.
For a power series \( \sum_{n=0}^{\infty} c_n(x-a)^n \), divergence occurs when the term \( c_n(x-a)^n \) causes the series sum to grow without bound or not settle to a specific value.In terms of radius:
For a power series \( \sum_{n=0}^{\infty} c_n(x-a)^n \), divergence occurs when the term \( c_n(x-a)^n \) causes the series sum to grow without bound or not settle to a specific value.In terms of radius:
- If the absolute value of \( x-a \) is greater than the radius of convergence \( R \), the series will diverge.
- If \( x-a \) is exactly equal to \( R \), the behavior of the series must be checked case by case.
Interval of Convergence
The interval of convergence is the range of \( x \) values for which a power series converges. It is crucial because it tells us where a power series is valid for representing a function.
For a power series centered at \( a \), the interval is generally written as:
For a power series centered at \( a \), the interval is generally written as:
- \( (a-R, a+R) \)
- At these points, the series may converge or diverge, depending on the specific series.
- Techniques such as substitution or specific tests (e.g., Ratio Test) can determine behavior at endpoints.
Other exercises in this chapter
Problem 30
Suppose you know that \( f^{(n)}(4) = \frac {( - 1)^n n!}{3^n(n + 1)} \) and the Taylor series of \( f \) centered at 4 converges to \( f(x) \) for all \( x \)
View solution Problem 30
Use a power series to approximate the definite integral to six decimal places. \( \int^{1/2}_0 \arctan (x/2) dx \)
View solution Problem 30
Test the series for convergence or divergence. \( \displaystyle \sum_{j = 1}^{\infty} ( - 1)^j \frac {\sqrt{j}}{j + 5} \)
View solution Problem 30
Use the Root Test to determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 0}^{\infty} \) (arctan \( n)^n \)
View solution