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Chapter 9: Problem 6

Find the interval of convergence of the given series. $$ \sum_{n=1}^{\infty} \frac{2}{3^{n+1} n^{2}} x^{n} $$

Short Answer

Expert verified
The interval of convergence is [-3, 3].

Step by step solution

01

Identify the Type of Series

The series is a power series of the form \( \sum_{n=1}^{\infty} a_n x^n \). In this exercise, \( a_n = \frac{2}{3^{n+1} n^{2}} \). We want to find the interval of convergence for this power series.
02

Set Up the Ratio Test

To find the interval of convergence, we will use the Ratio Test. For a series \( \sum_{n=1}^{\infty} a_n x^n \), the Ratio Test uses the absolute value \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \cdot |x| \lt 1 \). We want to find the values of \( x \) for which this limit is less than 1.
03

Calculate the Ratio of Consecutive Terms

Calculate the ratio \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{2}{3^{n+2}(n+1)^2}}{\frac{2}{3^{n+1}n^2}} \right| = \frac{n^2}{3(n+1)^2} \). This simplifies to \( \frac{n^2}{3n^2 + 6n + 3} \).
04

Compute the Limit for the Ratio Test

Now calculate the limit \( \lim_{n \to \infty} \frac{n^2}{3n^2 + 6n + 3} = \lim_{n \to \infty} \frac{1}{3 + \frac{6}{n} + \frac{3}{n^2}} = \frac{1}{3} \).
05

Apply the Ratio Test Result

According to the Ratio Test, the series converges if \( |x| \cdot \frac{1}{3} < 1 \), or \( |x| < 3 \). Therefore, the series converges for \( -3 < x < 3 \).
06

Check Endpoints for Convergence

The Ratio Test is inconclusive at the endpoints \( x = -3 \) and \( x = 3 \). Test these endpoints separately. For \( x = 3 \), the series becomes \( \sum_{n=1}^{\infty} \frac{2}{3n^2} \), which converges because it is a p-series with a power greater than 1. For \( x = -3 \), the series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^n \, 2}{3n^2} \), which also converges by the Alternating Series Test. Thus, the series converges at both endpoints.
07

Conclude the Interval of Convergence

After evaluating the endpoints, we conclude that the interval of convergence is \( [-3, 3] \), including both endpoints.

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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 useful tool for determining the convergence of an infinite series. It's particularly effective when dealing with power series. The test works by examining the limit of the absolute ratio between consecutive terms of the series. This helps us understand the behavior of the series as the number of terms approaches infinity.
To apply the Ratio Test, you calculate:
  • Determine the ratio of consecutive terms, i.e., \( \left| \frac{a_{n+1}}{a_n} \right| \).
  • Take the limit as \( n \) approaches infinity: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
  • Multiply this limit by \(|x|\) if you are working with a power series of the form \( \sum_{n=1}^{\infty} a_n x^n \).
The series converges if this result is less than 1. If it equals 1, the test is inconclusive, and additional methods must be employed to determine convergence.
Power Series
A power series is a series of the form \( \sum_{n=0}^{\infty} c_n x^n \), where each term of the series is a power of \( x \). This is central in calculus and analysis because it allows functions to be represented in a form amenable to analysis and computation.
  • The coefficients \( c_n \) dictate the behavior of the series.
  • The variable \( x \) can influence convergence, making the examination of the interval of convergence vital.
  • Finding this interval involves techniques such as the Ratio Test to ensure the series does not diverge at the extremes of \( x \).
In practice, power series can represent common functions, like exponential or trigonometric functions, within a certain distance to a center point \( x=0 \), known as the radius of convergence.
Alternating Series Test
The Alternating Series Test helps determine the convergence of series where the terms alternate in sign. Such series take the form \( \sum_{n=1}^{\infty} (-1)^n a_n \), introducing oscillating behavior in the series.
This test asserts that an alternating series converges if:
  • The absolute value of the terms \( a_n \) decreases monotonically; that is, each term is smaller than the preceding term.
  • The limit of \( a_n \) as \( n \) goes to infinity is zero.
If these conditions are met, the series converges. In the case of our exercise, the test ensures the series converges at the endpoint \( x = -3 \) by showing decrement in term size and reaching zero.
p-Series
A p-Series is a type of series of the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), where \( p \) is a constant exponent of \( n \). It is a fundamental series in mathematical analysis, and understanding its convergence is crucial.
  • The series converges if \( p > 1 \).
  • It diverges if \( p \leq 1 \).
For example, the series at the endpoint \( x = 3 \) transforms into a p-Series \( \sum_{n=1}^{\infty} \frac{2}{3n^2} \). Here, since \( p = 2 \) (greater than 1), the series converges. This gives us confidence in including the endpoint \( x = 3 \) in the interval of convergence.

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Most popular questions from this chapter

Approximate the sum of the given series with an error less than \(0.001\). $$ \sum_{n=2}^{\infty}(-1)^{n+1} \frac{1}{1+n+6 n^{2}} $$

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