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Magazine Chapter 9: Problem 35
Find the convergence set for each series. (a) \(\sum_{n=1}^{\infty} \frac{(3 x+1)^{n}}{n \cdot 2^{n}}\) (b) \(\sum_{n=1}^{\infty}(-1)^{n} \frac{(2 x-3)^{n}}{4^{n} \sqrt{n}}\)
Short Answer
Expert verified
(a) Converges on \((-\frac{4}{3}, \frac{1}{3})\), (b) Converges on \((1, \frac{7}{2})\)."
Step by step solution
01
Analyze Series (a)
The series given is \( \sum_{n=1}^{\infty} \frac{(3x+1)^{n}}{n \cdot 2^{n}} \). This is a power series of the form \( \sum a_n (x-c)^n \), where \( a_n = \frac{1}{n} \) and \( (x-c)^n = (3x+1)^n \). The series can be rewritten as \( \sum_{n=1}^{\infty} \left(\frac{3x+1}{2}\right)^n \cdot \frac{1}{n} \).
02
Use Ratio Test for Series (a)
To find the convergence set, apply the ratio test. For ratio test, examine \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Here, \( a_n = \frac{1}{n} \left(\frac{3x+1}{2}\right)^n \), making \( a_{n+1} = \frac{1}{n+1} \left(\frac{3x+1}{2}\right)^{n+1} \). So, the ratio is \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{3x+1}{2} \right| \cdot \frac{n}{n+1} \). As \( n \to \infty \), the limit becomes \( \left| \frac{3x+1}{2} \right| \). For the series to converge, this must be less than 1.
03
Solve Inequality for Series (a)
Solve \( \left| \frac{3x+1}{2} \right| < 1 \) to find the interval of convergence. \( \frac{-1}{2} < \frac{3x+1}{2} < 1 \). By solving, we get \(-3 < 3x+1 < 2\). Then, solve for \(x\): \( -4 < 3x < 1 \) and dividing by 3: \( \frac{-4}{3} < x < \frac{1}{3} \). So, the interval of convergence is \( \left(\frac{-4}{3}, \frac{1}{3}\right) \).
04
Analyze Series (b)
The series is \( \sum_{n=1}^{\infty} (-1)^{n} \frac{(2x-3)^{n}}{4^{n} \sqrt{n}} \). This is an alternating series where \( (x-c)^n = (2x-3)^n \). Rewriting the series gives \( \sum_{n=1}^{\infty} (-1)^{n} \frac{1}{\sqrt{n}} \left( \frac{2x-3}{4} \right)^n \).
05
Use Root Test for Series (b)
We'll use the root test to establish the radius of convergence. The test evaluates \( \lim_{n \to \infty} \sqrt[n]{\left| a_n \right|} \). Here, \( a_n = \frac{1}{\sqrt{n}} \left( \frac{2x-3}{4} \right)^n \). The root test analyzes \( \left| \frac{2x-3}{4} \right| \) because \( \lim_{n \to \infty} \sqrt[n]{\frac{1}{\sqrt{n}}} = 1 \). For convergence, this factor must be less than 1.
06
Solve Inequality for Series (b)
Solve \( \left| \frac{2x-3}{4} \right| < 1 \); hence, \( \frac{-1}{4} < \frac{2x-3}{4} < 1 \). By solving, \(-1 < 2x-3 < 4\). Rearrange to solve for \(x\): \(2 < 2x < 7\), which simplifies to \(1 < x < \frac{7}{2} \). The interval of convergence is \( (1, \frac{7}{2}) \).
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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 of the series, and \( c \) is the center of the series. Power series are infinite series that depend on a variable \( x \), and their terms are powers of \( (x-c) \). These are a cornerstone in calculus because they provide a way to approximate functions over an interval.
To effectively work with power series, you should understand:
To effectively work with power series, you should understand:
- The center \( c \): This is the point around which the series is expanded. Changing \( c \) impacts the interval within which the series converges.
- The coefficients \( a_n \): These determine the weight of each term in the series and are often determined through differentiation and evaluation of the function being approximated.
Ratio Test
The ratio test is a method for determining the absolute convergence of a series. It states that for a series \( \sum a_n \), if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \), then:
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \) or \( L = \infty \), the series diverges.
- If \( L = 1 \), the test is inconclusive.
Root Test
The root test is another tool to determine the convergence of a series. Similar to the ratio test, it offers a criterion stated as follows: for a series \( \sum a_n \), if \( \lim_{n \to \infty} \sqrt[n]{|a_n|} = L \), then:
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \), the series diverges.
- If \( L = 1 \), the test is inconclusive.
Interval of Convergence
The interval of convergence refers to the set of \( x \)-values for which a series converges. Finding this interval is crucial because it tells us the domain where the power series behaves well and closely approximates functions.
Here's how it's approached:
Here's how it's approached:
- Convergent at endpoints: This often requires separate testing of the series' behavior at the endpoints of the interval, as they may converge conditionally, diverge, or converge absolutely.
- Solving inequalities: As seen in both series (a) and (b), after applying the ratio or root test, solving the resulting inequality gives the open interval of convergence.
