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Chapter 6: Problem 15

Find the radius of convergence \(R\) and interval of convergence for \(\sum a_{n} x^{n}\) with the given coefficients \(a_{n}\). $$ \sum_{n=1}^{\infty} \frac{n x^{n}}{2^{n}} $$

Short Answer

Expert verified
The radius of convergence is 2, and the interval is \((-2, 2)\).

Step by step solution

01

Identify the series format

The given series is \( \sum_{n=1}^{\infty} \frac{n x^{n}}{2^{n}} \). This is a power series of the form \( \sum a_{n} x^{n} \) where \( a_{n} = \frac{n}{2^n} \).
02

Apply the Ratio Test

To find the radius of convergence \( R \), apply the Ratio Test:\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \].For our series, this becomes:\[ \lim_{n \to \infty} \left| \frac{\frac{n+1}{2^{n+1}}}{\frac{n}{2^n}} \right| = \lim_{n \to \infty} \left| \frac{n+1}{n} \cdot \frac{1}{2} \right| = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right) \cdot \frac{1}{2} \].
03

Evaluate the Limit

To evaluate the limit, simplify to:\[ \lim_{n \to \infty} \left( \frac{n+1}{2n} \right) = \lim_{n \to \infty} \left( \frac{1 + \frac{1}{n}}{2} \right) \].As \( n \to \infty \), \( \frac{1}{n} \to 0 \), so the limit is \( \frac{1}{2} \).
04

Find the Radius of Convergence

The Ratio Test implies convergence when the limit is less than 1:\[ \left| x \right| \cdot \frac{1}{2} < 1 \].Therefore, \( \left| x \right| < 2 \). This gives the radius of convergence \( R = 2 \).
05

Determine the Interval of Convergence

With the radius \( R = 2 \), the interval is initially \( -2 < x < 2 \). Check endpoints separately as the Ratio Test does not provide information on convergence at the endpoints.- For \( x = 2 \): the series becomes \( \sum \frac{n \cdot 2^n}{2^n} = \sum n \), which diverges.- For \( x = -2 \): the series becomes \( \sum \frac{n \cdot (-2)^n}{2^n} = \sum (-1)^n n \), which also diverges as it is not absolutely convergent.Thus, the interval of convergence is \( (-2, 2) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Interval of Convergence
The interval of convergence is a range of values where a power series converges. In a power series like \( \sum a_{n} x^n \), the series may not converge for all values of \( x \). Rather, it converges within an interval on the real number line.

To find this interval, we first find the radius of convergence \( R \), which determines how "wide" the convergence region is around the center of the series, typically \( x = 0 \). The series converges for \( |x| < R \). This results in an initial interval \( (-R, R) \).
  • We also need to evaluate the behavior at the endpoints, \(-R\) and \(R\), separately, as convergence can change at these points.
  • If the series converges at an endpoint, that endpoint is included in the interval. If not, it remains excluded.
  • For the given exercise, the radius of convergence is \( R = 2 \), and the interval is \((-2, 2)\), with both endpoints excluded because the series diverges there.
Power Series
A power series is an infinite series in the form \( \sum a_{n} x^n \), where \( a_{n} \) are coefficients and \( x \) is a variable. Each term in the series involves a power of \( x \), hence the name "power series."
  • These series are central in mathematical analysis, including calculus and functions.
  • They represent functions as infinite polynomials, allowing us to study function behaviors using series concepts.
  • Within its interval of convergence, a power series behaves like a polynomial.
The given exercise is a power series where \( a_{n} = \frac{n}{2^n} \). The series starts at \( n = 1 \) and continues indefinitely. Each coefficient plays a role in determining both the radius and interval of convergence.
Ratio Test
The Ratio Test is a tool for determining series convergence. It is particularly useful for power series. The test involves taking the limit of the absolute value of the ratio of successive terms.

For a series \( \sum a_{n} \), the Ratio Test is applied by evaluating:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}} \right|\]Results interpretation:
  • If the limit \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \), the series diverges.
  • If \( L = 1 \), the test is inconclusive.
In the provided exercise, the Ratio Test helps find that the series converges for values of \( x \) where \( |x| < 2 \), setting a foundation for determining the interval of convergence.
Convergence
Convergence in the context of series, especially power series, refers to the behavior of a series as its terms are added indefinitely. For a series to converge, the sums must approach a finite limit.
  • Absolute convergence implies convergence even when all terms are replaced by their absolute values.
  • Conditional convergence occurs when the series converges but not absolutely.
  • Divergence occurs when the series does not approach a finite limit.
For power series, convergence is generally analyzed within its interval of convergence. At specific points, like the endpoints of the interval, the series can behave differently. In the exercise, evaluating convergence showed that at the endpoints \( x = -2 \) and \( x = 2 \), the series diverges, confirming the interval as \( (-2, 2) \).

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Most popular questions from this chapter

The error in approximating the integral \(\int_{a}^{b} f(t) d t\) by that of a Taylor approximation \(\int_{a}^{b} P_{n}(t) d t\) is at most \(\int_{a}^{b} R_{n}(t) d t .\) The Taylor remainder estimate \(R_{n} \leq \frac{M}{(n+1) !}|x-a|^{n+1}\) guarantees that the integral of the Taylor polynomial of the given order approximates the integral of \(f\) with an error less than \(\frac{1}{10}\). a. Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than \(\frac{1}{100}\). b. Compare the accuracy of the polynomial integral estimate with the remainder estimate. \(\int_{0}^{\pi} \frac{\sin t}{t} d t ; P_{s}=1-\frac{x^{2}}{3 !}+\frac{x^{4}}{5 !}-\frac{x^{6}}{7 !}+\frac{x^{8}}{9 !}\) (You may assume that the absolute value of the ninth derivative of \(\frac{\sin t}{t}\) is bounded by \(0.1 .\) )

Use the binomial approximation \(\sqrt{1-x} \approx 1-\frac{x}{2}-\frac{x^{2}}{8}-\frac{x^{3}}{16}-\frac{5 x^{4}}{128}-\frac{7 x^{5}}{256}\) for \(|x|<1\) to approximate each number. Compare this value to the value given by a scientific calculator. \(\frac{1}{\sqrt{2}}\) using \(x=\frac{1}{2}\) in \((1-x)^{1 / 2}\)

Use the binomial approximation \(\sqrt{1-x} \approx 1-\frac{x}{2}-\frac{x^{2}}{8}-\frac{x^{3}}{16}-\frac{5 x^{4}}{128}-\frac{7 x^{5}}{256}\) for \(|x|<1\) to approximate each number. Compare this value to the value given by a scientific calculator. $$ \sqrt{5}=5 \times \frac{1}{\sqrt{5}} \text { using } x=\frac{4}{5} \text { in }(1-x)^{1 / 2} $$

Use the substitution \((b+x)^{r}=(b+a)^{r}\left(1+\frac{x-a}{b+a}\right)^{r} \quad\) in the binomial expansion to find the Taylor series of each function with the given center. $$ \sqrt{x+2} \text { at } a=1 $$

Compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of \(f\). $$ f(x)=e^{x} \cos x $$
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