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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![### Problem Statement:
Given the series:
\[
\sum_{n=1}^{\infty} \frac{11^n}{n}
\]
find the ratio \(\left|\frac{a_{n+1}}{a_n}\right|\).
### Explanation:
The series provided is an infinite sum, starting from \(n=1\) to infinity. Each term of the series is given by \(\frac{11^n}{n}\), where \(n\) is the summation variable. The goal is to determine the absolute value of the ratio of consecutive terms within this series.
### Steps:
1. Identify the general term \(a_n\) from the given series:
\[
a_n = \frac{11^n}{n}
\]
2. Write the next term \(a_{n+1}\):
\[
a_{n+1} = \frac{11^{n+1}}{n+1}
\]
3. Compute the ratio \(\left|\frac{a_{n+1}}{a_n}\right|\):
\[
\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{\frac{11^{n+1}}{n+1}}{\frac{11^n}{n}}\right| = \left|\frac{11^{n+1} \cdot n}{11^n \cdot (n+1)}\right| = \left|\frac{11 \cdot 11^n \cdot n}{11^n \cdot (n+1)}\right| = \left|\frac{11n}{n+1}\right|
\]
Thus, the ratio \(\left|\frac{a_{n+1}}{a_n}\right|\) simplifies to:
\[
\left|\frac{11n}{n+1}\right|
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1542e5de-e392-4bdb-9d95-e5abdf8e6267%2F6a54dce6-6827-494b-a923-92a1d80f9e39%2Fhtvrqz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Given the series:
\[
\sum_{n=1}^{\infty} \frac{11^n}{n}
\]
find the ratio \(\left|\frac{a_{n+1}}{a_n}\right|\).
### Explanation:
The series provided is an infinite sum, starting from \(n=1\) to infinity. Each term of the series is given by \(\frac{11^n}{n}\), where \(n\) is the summation variable. The goal is to determine the absolute value of the ratio of consecutive terms within this series.
### Steps:
1. Identify the general term \(a_n\) from the given series:
\[
a_n = \frac{11^n}{n}
\]
2. Write the next term \(a_{n+1}\):
\[
a_{n+1} = \frac{11^{n+1}}{n+1}
\]
3. Compute the ratio \(\left|\frac{a_{n+1}}{a_n}\right|\):
\[
\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{\frac{11^{n+1}}{n+1}}{\frac{11^n}{n}}\right| = \left|\frac{11^{n+1} \cdot n}{11^n \cdot (n+1)}\right| = \left|\frac{11 \cdot 11^n \cdot n}{11^n \cdot (n+1)}\right| = \left|\frac{11n}{n+1}\right|
\]
Thus, the ratio \(\left|\frac{a_{n+1}}{a_n}\right|\) simplifies to:
\[
\left|\frac{11n}{n+1}\right|
\]
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