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Chapter 9: Problem 29

Find the interval of convergence. $$\sum_{n=1}^{\infty} \frac{n^{2} x^{2 n}}{2^{2 n}}$$

Short Answer

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

Step by step solution

01

Write down the series

The given series is \( \sum_{n=1}^{\infty} \frac{n^{2} x^{2n}}{2^{2n}} \). This is a power series centered at 0 with terms \( a_n = \frac{n^2 x^{2n}}{2^{2n}} \).
02

Apply the Ratio Test

To find the interval of convergence, we use the Ratio Test. For the general term \( a_n = \frac{n^2 x^{2n}}{2^{2n}} \), consider the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \):\[\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{((n+1)^2 x^{2(n+1)}) / 2^{2(n+1)}}{(n^2 x^{2n}) / 2^{2n}} \right| = \left| \frac{(n+1)^2 x^2}{4 n^2} \right|.\]Simplify to get:\[\left| \frac{(n+1)^2}{n^2} \right| \cdot \left| x^2 \right| \cdot \frac{1}{4}.\]
03

Evaluate the limit as n approaches infinity

Evaluate the limit:\[ \lim_{n \to \infty} \left| \frac{(n+1)^2}{n^2} \right| = \lim_{n \to \infty} \left(1 + \frac{2}{n} + \frac{1}{n^2}\right) = 1.\] Thus, substituting back, we have:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{|x^2|}{4} = \frac{x^2}{4}.\]
04

Find interval from the limit result

For convergence, the Ratio Test requires this limit to be less than 1:\[\frac{x^2}{4} < 1.\]Multiply both sides by 4:\[x^2 < 4.\]Take the square root of both sides:\[|x| < 2.\]
05

Determine the interval bounds

The inequality \(|x| < 2\) gives the interval of convergence \((-2, 2)\). Check the endpoints to determine if they are included.For \(x = 2\) and \(x = -2\), substitute into the series. Both result in a series that diverges by the comparison test (compared to the p-series). Thus, the endpoints are not included in the interval.

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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 type of series involving terms of a variable raised to a power. Think of it like a polynomial, but one that can have infinitely many terms. The general form of a power series is:
  • \[ \sum_{n=0}^{\infty} c_n (x-a)^n \]
where:
  • \( c_n \) are the coefficients of the series,
  • \( x \) is the variable,
  • \( a \) is the center or the point around which the series is centered.
For our specific series, the terms are \( \frac{n^2 x^{2n}}{2^{2n}} \) and it is centered at zero. The power series starts at \( n = 1 \) and involves powers of \( x^2 \), reflecting its nature as a series that extends infinitely. Power series are often used to represent functions over a particular interval, known as the interval of convergence.
Ratio Test
The Ratio Test is a useful tool to determine if an infinite series converges. It is particularly handy for power series. Here's the basic idea:
  • For a given series \( \sum a_n \), calculate the ratio \( \left| \frac{a_{n+1}}{a_n} \right| \).
  • Check the limit \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
  • If this limit \(< 1\), the series converges.
  • If it is \(> 1\) or \(\infty\), the series diverges.
  • If it equals \(1\), the test is inconclusive.
In the exercise, the Ratio Test was applied to find the interval of convergence. By taking \( \left| \frac{(n+1)^2 x^2}{4 n^2} \right| \), simplifying, and evaluating the limit, we see that for the series to converge, \( \frac{x^2}{4} < 1 \). This inequality helps define the interval on which the series converges.
Limit Evaluation
In the context of series, limits help us understand the behavior of terms as \( n \) becomes very large. Evaluating limits is crucial when applying the Ratio Test. In our exercise, the limit of the ratio was evaluated as follows:
  • Calculate \( \lim_{n \to \infty} \left| \frac{(n+1)^2}{n^2} \right| \).
  • Use algebra to express \((n+1)^2 = n^2 + 2n + 1 \).
  • Divide each term by \( n^2 \) to get \( 1 + \frac{2}{n} + \frac{1}{n^2} \).
  • As \( n \to \infty \), terms \( \frac{2}{n} \) and \( \frac{1}{n^2} \) become \(0\).
  • Thus, the limit is \(1\).
This calculated limit played a key role in concluding that \( \frac{|x^2|}{4} < 1 \) is necessary for convergence of the series.
Infinite Series
Infinite series can look intimidating, but they're fundamentally a sequence of summed terms that go on forever. This concept underpins much of calculus and analysis in mathematics. Understanding infinite series involves:
  • Recognizing the series' structure: terms, general formula, and starting point.
  • Determining convergence: figuring out if the series totals to a finite number.
  • Using tests like the Ratio Test to evaluate convergence.
  • Finding the interval of convergence for power series, like \( (-2, 2) \) in our series.
Infinite series are applied in various fields like signal processing, quantum physics, and in solving differential equations. They allow us to approximate complex functions with simple terms for specific applications.

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Most popular questions from this chapter

A ball is dropped from a height of 10 feet and bounces. Each bounce is \(\frac{3}{4}\) of the height of the bounce before. Thus, after the ball hits the floor for the first time, the ball rises to a height of \(10\left(\frac{3}{4}\right)=7.5\) feet, and after it hits the floor for the second time, it rises to a height of \(7.5\left(\frac{3}{4}\right)=10\left(\frac{3}{4}\right)^{2}=5.625\) feet. (Assume that there is no air resistance.) (a) Find an expression for the height to which the ball rises after it hits the floor for the \(n^{\text {th }}\) time. (b) Find an expression for the total vertical distance the ball has traveled when it hits the floor for the first, second, third, and fourth times. (c) Find an expression for the total vertical distance the ball has traveled when it hits the floor for the \(n^{\text {th }}\) time. Express your answer in closed form.

Give an example of: A series \(\sum_{n=1}^{\infty} a_{n}\) that converges but \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) diverges.

Decide if the statements are true or false. Give an explanation for your answer. If all terms \(s_{n}\) of a sequence are less than a million, then the sequence is bounded.

True or false. Give an explanation for your answer. \(\sum C_{n}(x-1)^{n}\) and \(\sum C_{n} x^{n}\) have the same radius of convergence.

Which of the sequences I-IV is monotone and bounded for \(n \geq 1 ?\) I. \(s_{n}=10-\frac{1}{n}\) II. \(s_{n}=\frac{10 n+1}{n}\) III. \(s_{n}=\cos n\) IV. \(s_{n}=\ln n\) (a) I (b) I and II (c) II and IV (d) I, II, and III
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