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Magazine Chapter 11: Problem 18
Find the radius of convergence and interval of convergence of the series. $$\sum_{n=1}^{\infty} \frac{n}{4^{n}}(x+1)^{n}$$
Short Answer
Expert verified
The radius of convergence is 4, and the interval of convergence is \([-5, 3)\).
Step by step solution
01
Identify the Power Series Form
The given series can be expressed in a more standard form: \( \sum_{n=1}^{\infty} \frac{n}{4^{n}} (x+1)^n \). This is a power series centered at \( x = -1 \). The general term is \( a_n (x+1)^n \) where \( a_n = \frac{n}{4^n} \).
02
Use the Ratio Test to Determine Convergence
Apply the ratio test to find the radius of convergence. The ratio test states that for a series \( \sum a_n \), the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \) if less than 1, indicates convergence. Thus, we compute:\[ L = \lim_{n \to \infty} \left| \frac{(n+1)}{4^{n+1}} (x+1)^{n+1} \times \frac{4^n}{n} \times \frac{1}{(x+1)^n} \right| = \lim_{n \to \infty} \left| \frac{n+1}{4n} \right| |x+1|. \]
03
Simplify the Ratio and Find the Limit
Simplify the expression from the ratio test:\[ L = \frac{|x+1|}{4} \cdot \lim_{n \to \infty} \frac{n+1}{n} = \frac{|x+1|}{4} \cdot \lim_{n \to \infty} \left(1 + \frac{1}{n}\right) = \frac{|x+1|}{4}. \]
04
Determine the Radius of Convergence
For convergence, the ratio \( \frac{|x+1|}{4} < 1 \) must be satisfied. Solving for \( |x+1| \), we find that \(|x+1| < 4\). Hence, the radius of convergence \( R = 4 \).
05
Identify the Interval of Convergence
The interval of convergence is \(-5 < x < 3\), obtained from the inequality \(-4 < x+1 < 4\). We need to check the endpoints \( x = -5 \) and \( x = 3 \) separately to determine if the series converges at those points.
06
Check Convergence at the Endpoints
At \( x = -5 \), the series becomes \( \sum_{n=1}^{\infty} \frac{n(-1)^n}{4^n} \), an alternating series that converges by the alternating series test.At \( x = 3 \), the series becomes \( \sum_{n=1}^{\infty} \frac{n}{4^n} \cdot 4^n = \sum_{n=1}^{\infty} n \), a divergent series since the terms \( n \) do not tend to zero.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interval of Convergence
Understanding the interval of convergence is essential when dealing with power series. For a power series \( \sum a_n (x-c)^n \), the interval of convergence is the set of all \( x \) values for which the series converges.
This interval can be finite, semi-infinite, or infinite, depending on how far \( x \) can be from the center \( c \) while maintaining convergence.
To find the interval of convergence, we first determine the radius of convergence \( R \). The interval is then \( (c-R, c+R) \) before checking endpoints.
The endpoints need special consideration since convergence at endpoints can differ.
Using tests like the ratio test can help determine the interval of convergence quickly for standard cases. However, one must always verify convergence at endpoints separately, often with tests like the alternating series test or a comparison test.
This interval can be finite, semi-infinite, or infinite, depending on how far \( x \) can be from the center \( c \) while maintaining convergence.
To find the interval of convergence, we first determine the radius of convergence \( R \). The interval is then \( (c-R, c+R) \) before checking endpoints.
The endpoints need special consideration since convergence at endpoints can differ.
Using tests like the ratio test can help determine the interval of convergence quickly for standard cases. However, one must always verify convergence at endpoints separately, often with tests like the alternating series test or a comparison test.
Power Series
A power series is an infinite series of the form \( \sum a_n (x-c)^n \), where \( a_n \) are the coefficients and \( x-c \) terms are powers of \( x \) centered at \( c \).
This series acts like a polynomial of infinite degree and is useful because it can approximate many types of functions.
In our exercise, the power series in question is centered at \( x = -1 \). We rewrite it as \( \sum \frac{n}{4^n} (x+1)^n \), identifying \( c = -1 \) and \( a_n = \frac{n}{4^n} \).
The concept of a power series is crucial in calculus and complex analysis, providing a versatile tool for solution and approximation of functions.
This series acts like a polynomial of infinite degree and is useful because it can approximate many types of functions.
In our exercise, the power series in question is centered at \( x = -1 \). We rewrite it as \( \sum \frac{n}{4^n} (x+1)^n \), identifying \( c = -1 \) and \( a_n = \frac{n}{4^n} \).
The concept of a power series is crucial in calculus and complex analysis, providing a versatile tool for solution and approximation of functions.
- Series convergence depends on \( x \).
- It resembles polynomials but has potentially infinite terms.
- Different tests determine where it converges.
Ratio Test
The ratio test is a method used to determine the convergence of series and is especially popular for power series.
According to the ratio test for a series \( \sum a_n \), compute \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If \( L < 1 \), then the series converges.
In our example, simplifying the ratio gives us \( L = \frac{|x+1|}{4} \). By solving \( \frac{|x+1|}{4} < 1 \), we derive the condition \(|x+1| < 4\), identifying the radius of convergence \( R = 4 \).
The ratio test efficiently helps find where within the interval the series converges and is often a good first step for these kinds of problems.
According to the ratio test for a series \( \sum a_n \), compute \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If \( L < 1 \), then the series converges.
In our example, simplifying the ratio gives us \( L = \frac{|x+1|}{4} \). By solving \( \frac{|x+1|}{4} < 1 \), we derive the condition \(|x+1| < 4\), identifying the radius of convergence \( R = 4 \).
The ratio test efficiently helps find where within the interval the series converges and is often a good first step for these kinds of problems.
- Convergence: \( L < 1 \)
- Inconclusive: \( L = 1 \)
- Divergence: \( L > 1 \)
Alternating Series Test
The alternating series test is used specifically for series whose terms alternate in sign.
It provides a straightforward way to determine convergence when the series switches between positive and negative.
The test states that an alternating series \( \sum (-1)^n b_n \), converges if:
In the original exercise, at \( x = -5 \), the series becomes \( \sum \frac{n(-1)^n}{4^n} \), which passes the alternating series test and converges.
However, at \( x = 3 \), the series \( \sum n \) diverges because the terms \( n \) do not tend to zero.
This test provides an intuitive check for convergence by focusing on the nature of alternating sequences and their tendency toward zero.
It provides a straightforward way to determine convergence when the series switches between positive and negative.
The test states that an alternating series \( \sum (-1)^n b_n \), converges if:
- \( b_{n+1} \leq b_n \) for all \( n \), ensuring the sequence is decreasing.
- \( \lim_{n \to \infty} b_n = 0 \).
In the original exercise, at \( x = -5 \), the series becomes \( \sum \frac{n(-1)^n}{4^n} \), which passes the alternating series test and converges.
However, at \( x = 3 \), the series \( \sum n \) diverges because the terms \( n \) do not tend to zero.
This test provides an intuitive check for convergence by focusing on the nature of alternating sequences and their tendency toward zero.
