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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

Identify the Series Coefficients

The given series is \( \sum_{n=1}^{\infty} \frac{n^{2} x^{2n}}{2^{2n}} \). We can see that this is in the form of a power series \( \sum a_n x^n \) with \( a_n = \frac{n^{2}}{2^{2n}} \) and terms \( x^{2n} \).
02

Apply the Ratio Test

To find the interval of convergence, apply the ratio test to the general term \( a_n x^{2n} = \frac{n^2 x^{2n}}{2^{2n}} \). Calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1} x^{2(n+1)}}{a_n x^{2n}} \right| \).
03

Simplify the Ratio

The ratio \( \left| \frac{\frac{(n+1)^{2} x^{2(n+1)}}{2^{2(n+1)}}}{\frac{n^2 x^{2n}}{2^{2n}}} \right| = \left| \frac{(n+1)^2 x^{2} x^{2n}}{4n^2 x^{2n}} \right| = \left| \frac{(n+1)^2 x^2}{4n^2} \right| \).
04

Take the Limit of the Simplified Ratio

Calculate \( \lim_{n \to \infty} \left| \frac{(n+1)^2 x^2}{4n^2} \right| = |x^2| \cdot \lim_{n \to \infty} \left| \frac{(n+1)^2}{4n^2} \right| = |x^2| \cdot \frac{1}{4} = \frac{x^2}{4} \).
05

Determine Convergence from the Ratio Test

According to the ratio test, the series converges if \( \frac{x^2}{4} < 1 \). This simplifies to \( x^2 < 4 \) or \( -2 < x < 2 \).
06

Check Endpoints for Convergence

Test the endpoints \( x = -2 \) and \( x = 2 \) separately to see if they converge. Substitute \( x = 2 \) into the original series: \( \sum \frac{n^2 4^n}{4^n} = \sum n^2 \), which diverges. Substitute \( x = -2 \) gives \( \sum \frac{n^2 4^n}{4^n} = \sum n^2 \), which also diverges. Both endpoints do not contribute to the convergence interval.

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Key Concepts

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

Ratio Test
The ratio test is a powerful tool used to determine the convergence of an infinite series. It involves examining the ratio of successive terms in the series. For a given power series \( \sum a_n x^n \), we apply the ratio test by calculating the limit:
  • \( L = \lim_{n \to \infty} \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| \)
In simpler terms, you compare the \(n+1\) term to the \(n\) term of the series. If \( L < 1 \), the series converges. If \( L > 1 \), it diverges. And if \( L = 1 \), the test is inconclusive. In the given problem, the limit simplifies to \( \frac{x^2}{4} \), guiding the interval determination.
Power Series
A power series is an infinite series in the form \( \sum c_n x^n \), where \( c_n \) are the coefficients of the series, and \( x \) is a variable. This kind of series is of great importance in calculus because it allows functions to be expressed as infinite polynomials.
  • Power series are used to describe a wide range of functions.
  • They converge to a particular function within a certain interval called the interval of convergence.
In this exercise, the series \( \sum \frac{n^2 x^{2n}}{2^{2n}} \) translates to a power series with coefficients \( \frac{n^2}{2^{2n}} \), and the variable portions are \( x^{2n} \). Understanding this setup is crucial for applying tests like the ratio test.
Convergence
Convergence in the context of series analysis refers to whether the sum of an infinite series approaches a finite limit. For power series, this is only guaranteed within a specific range of \( x \) values known as the interval of convergence.
  • A convergent series implies that as you sum more terms, the total approaches a particular value.
  • Divergent series indicates that the sum does not settle on a number as more terms are added.
For the series in problem \( \sum \frac{n^2 x^{2n}}{2^{2n}} \), using the ratio test, the series converges if \( -2 < x < 2 \). This means that any \( x \) value within these bounds will result in the series summing to a finite number.
Endpoints
Checking endpoints is a crucial step in determining the complete interval of convergence. For the given series, after applying the ratio test, we arrived at an interval from \(-2\) to \(2\) but need to test these boundary values separately.
  • Substitute \( x = 2 \) into the original series and check if it converges. If the series sums to infinity or does not stabilize, it diverges.
  • Similarly, substitute \( x = -2 \) and analyze the behavior of the series for convergence or divergence.
In this exercise, both \( x = 2 \) and \( x = -2 \) lead to the series \( \sum n^2 \), which diverges. Thus, these endpoints do not contribute, resulting in an open interval (-2, 2) for convergence.

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Most popular questions from this chapter

Decide if the statements are true or false. Give an explanation for your answer.If \(\sum a_{n}\) is absolutely convergent, then it is convergent.

Explain what is wrong with the statement. If a convergent sequence consists entirely of terms greater than \(2,\) then the limit of the sequence must be greater than 2.

Show that if \(\sum\left|a_{n}\right|\) converges, then \(\sum(-1)^{n} a_{n}\) converges.

Decide if the statements are true or false. Give an explanation for your answer.If \(\sum\left|a_{n}\right|\) converges, then \(\sum(-1)^{n}\left|a_{n}\right|\) converges.

This problem defines the Calkin-Wilf-Newman sequence of positive rational numbers. The sequence is remarkable because every positive rational number appears as one of its terms and none appears more than once. Every real number \(x\) can be written as an integer \(A\) plus a number \(B\) where \(0 \leq B<1 .\) For example, for \(x=12 / 5=2+2 / 5\) we have \(A=2\) and \(B=2 / 5 .\) For \(x=3=3+0\) we have \(A=3\) and \(B=0 .\) Define the function \(f(x)\) by \(f(x)=A+(1-B)\) For example, \(f(12 / 5)=2+(1-2 / 5)=13 / 5\) and \(f(3)=3+(1-0)=4\) (a) Evaluate \(f(x)\) for \(x=25 / 8,13 / 9,\) and \(\pi\) (b) Find the first six terms of the recursively defined Calkin-Wilf-Newman sequence: \(s_{n}=1 / f\left(s_{n-1}\right)\) for \(n>1\) and \(s_{1}=1\)
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