Problem 15
Question
Find the radius of convergence and interval of convergence of the series. \( \sum_{n = 8}^{\infty} \frac {(x - 2)^n}{n^2 + 1} \)
Step-by-Step Solution
Verified Answer
The radius of convergence is 1, and the interval of convergence is
[1, 3].
1Step 1: Identify the series form
The given series is \( \sum_{n = 8}^{\infty} \frac{(x-2)^n}{n^2 + 1} \). It is of the form \( \sum a_n(x-c)^n \), where \( a_n = \frac{1}{n^2 + 1} \) and \( c = 2 \).
2Step 2: Apply the Ratio Test
To determine the radius of convergence, we apply the Ratio Test: \( \lim_{n \to \infty} \left| \frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n} \right| \). Substitute \( a_n = \frac{1}{n^2 + 1} \) to get \( \lim_{n \to \infty} \left| \frac{(x-2)^{n+1}}{(n+1)^2 + 1} \cdot \frac{n^2 + 1}{(x-2)^n} \right| = \lim_{n \to \infty} \left| \frac{(x-2)(n^2 + 1)}{(n+1)^2 + 1} \right| \).
3Step 3: Simplify the Ratio Test expression
Simplify the expression: \( \lim_{n \to \infty} \left| (x-2) \cdot \frac{n^2 + 1}{(n+1)^2 + 1} \right| \). This further reduces to \( \lim_{n \to \infty} \left| (x-2) \cdot \frac{1 + \frac{1}{n^2}}{1 + \frac{2}{n} + \frac{2}{n^2}} \right| \). As \( n \to \infty \), the ratio becomes \( |x-2| \).
4Step 4: Find the radius of convergence
Since the series converges when \( |x-2| < 1 \), the radius of convergence \( R \) is 1. Thus, the interval of convergence is \( (2-1, 2+1) = (1, 3) \).
5Step 5: Verify endpoints for interval of convergence
Test the endpoints \( x = 1 \) and \( x = 3 \):- For \( x = 1 \), the series becomes \( \sum_{n=8}^{\infty} \frac{(-1)^n}{n^2+1} \). This is convergent by the Alternating Series Test.- For \( x = 3 \), the series becomes \( \sum_{n=8}^{\infty} \frac{1}{n^2+1} \), which converges by the p-series test with \( p > 1 \).Thus, both endpoints are included in the interval of convergence.
Key Concepts
Interval of ConvergenceRatio TestAlternating Series TestP-Series Test
Interval of Convergence
The interval of convergence is a crucial concept when working with power series. It defines the set of all values of \( x \) for which the series converges. For the series \( \sum_{n = 8}^{\infty} \frac{(x-2)^n}{n^2 + 1} \), this interval is derived from the radius of convergence.
To find the interval, we identify the center of the series, \( c = 2 \), and apply the ratio test to get \( |x-2| < 1 \). This inequality means the series will converge for values of \( x \) between 1 and 3. It initially gives us the interval \((1, 3)\).
Besides just the interior points, we need to check the endpoints separately at \( x = 1 \) and \( x = 3 \) to see if the series converges at these points. Verification of convergence at these endpoints confirms that they should be included or excluded based on whether convergence occurs. In this problem, testing the endpoints shows that both are included, leading to the closed interval \([1, 3]\).
To find the interval, we identify the center of the series, \( c = 2 \), and apply the ratio test to get \( |x-2| < 1 \). This inequality means the series will converge for values of \( x \) between 1 and 3. It initially gives us the interval \((1, 3)\).
Besides just the interior points, we need to check the endpoints separately at \( x = 1 \) and \( x = 3 \) to see if the series converges at these points. Verification of convergence at these endpoints confirms that they should be included or excluded based on whether convergence occurs. In this problem, testing the endpoints shows that both are included, leading to the closed interval \([1, 3]\).
Ratio Test
The Ratio Test is a popular method for determining the radius of convergence of a power series. To use this test for our series \( \sum_{n = 8}^{\infty} \frac{(x-2)^n}{n^2 + 1} \), we set up the limit:
\[ l \lim_{n \to \infty} \frac{|a_{n+1} (x-c)^{n+1}|}{|a_n (x-c)^n|}. \]
Substituting the terms from our series gives:
\[ \lim_{n \to \infty} \left| \frac{(x-2) (n^2 + 1)}{(n+1)^2 + 1} \right|. \]
After simplifying, this limit equals \( |x-2| \).
A convergent series satisfies \(|x-2| < 1\), leading to a radius of convergence \( R = 1 \). The distance from the center \( c = 2 \) to the ends of the interval 1 and 3 is the radius of convergence, confirming the size of the interval from center point 2.
\[ l \lim_{n \to \infty} \frac{|a_{n+1} (x-c)^{n+1}|}{|a_n (x-c)^n|}. \]
Substituting the terms from our series gives:
\[ \lim_{n \to \infty} \left| \frac{(x-2) (n^2 + 1)}{(n+1)^2 + 1} \right|. \]
After simplifying, this limit equals \( |x-2| \).
A convergent series satisfies \(|x-2| < 1\), leading to a radius of convergence \( R = 1 \). The distance from the center \( c = 2 \) to the ends of the interval 1 and 3 is the radius of convergence, confirming the size of the interval from center point 2.
Alternating Series Test
The Alternating Series Test provides a method to determine the convergence of series where the terms alternate in sign, such as \( \sum_{n=8}^{\infty} \frac{(-1)^n}{n^2+1} \). This scenario occurs at the endpoint \( x = 1 \) of the interval.
The test states that if the magnitude of the terms \( a_n \) decreases monotonically and the limit of these terms is zero, the series converges.
For our series at \( x = 1 \), the terms are \( \frac{1}{n^2+1} \), which decrease as \( n \) gets larger and also approach zero.
Both criteria of the test are satisfied which confirms convergence of the series at \( x = 1 \). This guarantees that the endpoint 1 can be included in the interval of convergence.
The test states that if the magnitude of the terms \( a_n \) decreases monotonically and the limit of these terms is zero, the series converges.
For our series at \( x = 1 \), the terms are \( \frac{1}{n^2+1} \), which decrease as \( n \) gets larger and also approach zero.
Both criteria of the test are satisfied which confirms convergence of the series at \( x = 1 \). This guarantees that the endpoint 1 can be included in the interval of convergence.
P-Series Test
The P-Series Test is vital when dealing with series of the form \( \sum \frac{1}{n^p} \). It tells us that a series converges if \( p > 1 \). In our problem, this test is applied at the endpoint \( x = 3 \).
Substituting into the series, we have \( \sum_{n=8}^{\infty} \frac{1}{n^2+1} \). This resembles a p-series with \( p = 2 \).
Since 2 is greater than 1, by the P-Series Test, this series converges.
Hence, it indicates convergence at \( x = 3 \) and thus, allows the endpoint to be included in our interval of convergence \([1, 3]\).
Substituting into the series, we have \( \sum_{n=8}^{\infty} \frac{1}{n^2+1} \). This resembles a p-series with \( p = 2 \).
Since 2 is greater than 1, by the P-Series Test, this series converges.
Hence, it indicates convergence at \( x = 3 \) and thus, allows the endpoint to be included in our interval of convergence \([1, 3]\).
Other exercises in this chapter
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