Chapter 9: Problem 3
Find the convergence set for the given power series. $$ \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} $$
Short Answer
Expert verified
The convergence set is \(-1 \leq x \leq 1\).
Step by step solution
01
Identify the Type of Series
The given series is a power series, represented as \( \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \). Here, each term is of the form \( a_n x^n \) where \( a_n = \frac{1}{n^2} \). This is a real power series in \( x \).
02
Apply the Ratio Test
To find the radius of convergence, we use the Ratio Test. Consider two consecutive terms of the series: \( a_n x^n = \frac{x^n}{n^2} \) and \( a_{n+1} x^{n+1} = \frac{x^{n+1}}{(n+1)^2} \). The Ratio Test requires us to evaluate \( \lim_{n \to \infty} \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| \).
03
Simplify the Ratio Expression
Calculate the ratio: \[\left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| = \left| \frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n} \right| = \left| x \right| \cdot \left( \frac{n^2}{(n+1)^2} \right).\]This simplifies further to \( \left| x \right| \cdot \left( \frac{n^2}{(n+1)^2} \right) \).
04
Evaluate the Limit
Evaluate the limit as \( n \to \infty \):\[\lim_{n \to \infty} \left| x \right| \cdot \left( \frac{n^2}{(n+1)^2} \right) = \left| x \right| \cdot \lim_{n \to \infty} \left( \frac{n^2}{(n+1)^2} \right) = \left| x \right| \cdot 1 = \left| x \right|.\]According to the Ratio Test, the series converges if \( \left| x \right| < 1 \).
05
Determine the Convergence Set
From the evaluation of the Ratio Test, we conclude that the power series converges when \( \left| x \right| < 1 \), i.e., \( -1 < x < 1 \). For \( x = 1 \) or \( x = -1 \), the series should be checked separately. Using the Alternating Series Test or by considering the behavior of the series \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \) and \( \sum_{n=1}^{\infty} \frac{1}{n^2} \), both converge. Thus, the convergence set is \( -1 \leq x \leq 1 \).
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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 handy tool used to determine the convergence of a power series. It works by examining the ratio of consecutive terms in a series. In our exercise, we look at the series \( \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \) and focus on two consecutive terms: \( \frac{x^{n+1}}{(n+1)^2} \) and \( \frac{x^{n}}{n^2} \).
The Ratio Test requires us to find:
For the series to converge using the Ratio Test, this last expression must be less than 1. Hence,
The Ratio Test requires us to find:
- \( \left| \frac{a_{n+1} x^{n+1}}{a_n x^n} \right| \)
For the series to converge using the Ratio Test, this last expression must be less than 1. Hence,
- \( \left| x \right| < 1 \)
Radius of Convergence
The radius of convergence is a key aspect of a power series' behavior. It defines an interval around a central value in which the series converges.
In our series \( \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \), the variable \( x \) determines if the series will converge. After applying the Ratio Test, we determined that
In our series \( \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} \), the variable \( x \) determines if the series will converge. After applying the Ratio Test, we determined that
- \( \left| x \right| < 1 \)
Alternating Series Test
This test is used to check the convergence of series where terms alternate in sign. For endpoint evaluation in our power series, let’s consider \( x = -1 \).
The series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \), which is an alternating series. It meets the criteria:
The series becomes \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \), which is an alternating series. It meets the criteria:
- The terms \( \frac{1}{n^2} \) decrease steadily.
- The limit of these terms as \( n \to \infty \) is zero.
Limit Evaluation
Evaluating limits plays a pivotal role in ascertaining the convergence or divergence of a series. It is often used in conjunction with tests like the Ratio Test to define convergence properties.
As shown earlier, the limit:
This crucial step confirms the interval of convergence derived from the Ratio Test. Moreover, the limit ensures that the convergence is consistent across the interval \(-1 < x < 1\), establishing \( \left| x \right| < 1 \) as valid. Thus, our interval of convergence is bounded neatly by evaluating this limit.
As shown earlier, the limit:
- \( \lim_{n \to \infty} \left| x \right| \cdot \left( \frac{n^2}{(n+1)^2} \right) \)
This crucial step confirms the interval of convergence derived from the Ratio Test. Moreover, the limit ensures that the convergence is consistent across the interval \(-1 < x < 1\), establishing \( \left| x \right| < 1 \) as valid. Thus, our interval of convergence is bounded neatly by evaluating this limit.