Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Finding the Interval of Convergence for the Given Series
To determine where the given series converges, we need to analyze the following series:
\[
\sum_{n=0}^{\infty} (n+5)! (x + 12)^n
\]
**Step-by-Step Process to Find the Interval of Convergence:**
1. **Identify the General Term:**
\[
a_n = (n+5)!(x + 12)^n
\]
2. **Apply the Ratio Test:**
\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
\]
For our series:
\[
a_{n+1} = (n+6)! (x + 12)^{n+1}
\]
\[
\frac{a_{n+1}}{a_n} = \frac{(n+6)! (x + 12)^{n+1}}{(n+5)! (x + 12)^n} = (n+6) |x + 12|
\]
3. **Simplify:**
\[
\lim_{n \to \infty} (n+6)|x + 12| = \lim_{n \to \infty} n|x + 12|
\]
The series converges when:
\[
\lim_{n \to \infty} n|x + 12| < 1
\]
- If \( |x + 12| > 0 \), the limit approaches infinity, so the series diverges.
- If \( |x + 12| = 0 \) or \( x = -12 \), the limit is 0, leading to convergence.
4. **Conclusion:**
- The series converges only at \( x = -12 \).
Thus, the interval of convergence is just the single point \( \{ x = -12 \} \).
### Visualization and Explanation:
There are no graphs or diagrams provided in the image. However, if we were to provide a graphical explanation, it would include an illustration of the function's convergence point on the x-axis, indicating that the series converges only at \( x = -12 \), corresponding to the derived](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1542e5de-e392-4bdb-9d95-e5abdf8e6267%2Fba77aca8-56bc-46a4-88a6-e6f00c64a991%2Fpf1vfyt_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Interval of Convergence for the Given Series
To determine where the given series converges, we need to analyze the following series:
\[
\sum_{n=0}^{\infty} (n+5)! (x + 12)^n
\]
**Step-by-Step Process to Find the Interval of Convergence:**
1. **Identify the General Term:**
\[
a_n = (n+5)!(x + 12)^n
\]
2. **Apply the Ratio Test:**
\[
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|
\]
For our series:
\[
a_{n+1} = (n+6)! (x + 12)^{n+1}
\]
\[
\frac{a_{n+1}}{a_n} = \frac{(n+6)! (x + 12)^{n+1}}{(n+5)! (x + 12)^n} = (n+6) |x + 12|
\]
3. **Simplify:**
\[
\lim_{n \to \infty} (n+6)|x + 12| = \lim_{n \to \infty} n|x + 12|
\]
The series converges when:
\[
\lim_{n \to \infty} n|x + 12| < 1
\]
- If \( |x + 12| > 0 \), the limit approaches infinity, so the series diverges.
- If \( |x + 12| = 0 \) or \( x = -12 \), the limit is 0, leading to convergence.
4. **Conclusion:**
- The series converges only at \( x = -12 \).
Thus, the interval of convergence is just the single point \( \{ x = -12 \} \).
### Visualization and Explanation:
There are no graphs or diagrams provided in the image. However, if we were to provide a graphical explanation, it would include an illustration of the function's convergence point on the x-axis, indicating that the series converges only at \( x = -12 \), corresponding to the derived
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