Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.3: Geometric Sequences
Problem 49E
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Question
![### Interval of Convergence
To determine the interval of convergence for the given series:
\[ \sum_{n=2}^{\infty} \frac{(x - 9)^n}{\ln(9n)} \]
we must analyze the series and apply convergence tests such as the Ratio Test or the Root Test.
#### Steps to Solve:
1. **Define the General Term of the Series:** Identify the general term \( a_n \) from the series:
\[
a_n = \frac{(x - 9)^n}{\ln(9n)}
\]
2. **Apply the Ratio Test:** The Ratio Test is often used to determine the radius of convergence for power series. According to the Ratio Test, compute:
\[
L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
\]
Here,
\[
a_{n+1} = \frac{(x - 9)^{n+1}}{\ln[9(n + 1)]}
\]
Therefore,
\[
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x - 9)^{n+1}}{\ln[9(n + 1)]} \times \frac{\ln(9n)}{(x - 9)^n} \right| = \left| (x - 9) \frac{\ln(9n)}{\ln[9(n + 1)]} \right|
\]
3. **Simplify and Find the Limit:** Simplify the expression within the limit and evaluate as \( n \) approaches infinity:
\[
L = \lim_{n \to \infty} \left| (x - 9) \frac{\ln(9n)}{\ln[9(n + 1)]} \right|
\]
Notice that as \( n \to \infty \), \( \ln[9(n + 1)] \) behaves similarly to \( \ln(9n) \). Thus, the fraction approaches 1, yielding:
\[
L = |x - 9|
\]
4. **Determine the Radius of Convergence:** According to the Ratio Test, the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1542e5de-e392-4bdb-9d95-e5abdf8e6267%2F805cd2b4-f2b5-4cfb-b2ff-a31904fe701b%2Fynqog5e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Interval of Convergence
To determine the interval of convergence for the given series:
\[ \sum_{n=2}^{\infty} \frac{(x - 9)^n}{\ln(9n)} \]
we must analyze the series and apply convergence tests such as the Ratio Test or the Root Test.
#### Steps to Solve:
1. **Define the General Term of the Series:** Identify the general term \( a_n \) from the series:
\[
a_n = \frac{(x - 9)^n}{\ln(9n)}
\]
2. **Apply the Ratio Test:** The Ratio Test is often used to determine the radius of convergence for power series. According to the Ratio Test, compute:
\[
L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
\]
Here,
\[
a_{n+1} = \frac{(x - 9)^{n+1}}{\ln[9(n + 1)]}
\]
Therefore,
\[
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x - 9)^{n+1}}{\ln[9(n + 1)]} \times \frac{\ln(9n)}{(x - 9)^n} \right| = \left| (x - 9) \frac{\ln(9n)}{\ln[9(n + 1)]} \right|
\]
3. **Simplify and Find the Limit:** Simplify the expression within the limit and evaluate as \( n \) approaches infinity:
\[
L = \lim_{n \to \infty} \left| (x - 9) \frac{\ln(9n)}{\ln[9(n + 1)]} \right|
\]
Notice that as \( n \to \infty \), \( \ln[9(n + 1)] \) behaves similarly to \( \ln(9n) \). Thus, the fraction approaches 1, yielding:
\[
L = |x - 9|
\]
4. **Determine the Radius of Convergence:** According to the Ratio Test, the
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