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:**
Write \( S = \sum_{n=5}^{\infty} \frac{1}{n(n-1)} \) as a telescoping series and find its sum.
**Solution Steps:**
1. **Identify the Series:**
The given series is \( S = \sum_{n=5}^{\infty} \frac{1}{n(n-1)} \).
2. **Decompose into Partial Fractions:**
The goal is to rewrite \( \frac{1}{n(n-1)} \) as a difference of two fractions:
\[
\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}
\]
3. **Form the Telescoping Series:**
Substitute back into the series:
\[
S = \sum_{n=5}^{\infty} \left( \frac{1}{n-1} - \frac{1}{n} \right)
\]
This type of series allows terms to cancel out sequentially (telescoping effect).
4. **Calculate the Partial Sum \( S_N \):**
The partial sum \( S_N \), from \( n=5 \) to \( n=N \), is
\[
S_N = \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \ldots + \left( \frac{1}{N-1} - \frac{1}{N} \right)
\]
5. **Show Cancellation:**
Observe that this series telescopes such that most terms cancel:
\[
S_N = \frac{1}{4} - \frac{1}{N}
\]
6. **Calculate the Infinite Sum \( S \):**
As \( N \to \infty \), \( \frac{1}{N} \to 0 \), so the sum of the infinite series is:
\[
S = \frac{1}{4}
\]
**Solutions:**
- \( S_N = \frac{1}{4} - \frac{1}{N} \)
- \( S = \frac{1}{4} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2daf8f3-113d-48f5-9b15-0b37d7046821%2Fa0b1f38c-e30b-4484-b00f-562a4681f33e%2Fv18o73r_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Write \( S = \sum_{n=5}^{\infty} \frac{1}{n(n-1)} \) as a telescoping series and find its sum.
**Solution Steps:**
1. **Identify the Series:**
The given series is \( S = \sum_{n=5}^{\infty} \frac{1}{n(n-1)} \).
2. **Decompose into Partial Fractions:**
The goal is to rewrite \( \frac{1}{n(n-1)} \) as a difference of two fractions:
\[
\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}
\]
3. **Form the Telescoping Series:**
Substitute back into the series:
\[
S = \sum_{n=5}^{\infty} \left( \frac{1}{n-1} - \frac{1}{n} \right)
\]
This type of series allows terms to cancel out sequentially (telescoping effect).
4. **Calculate the Partial Sum \( S_N \):**
The partial sum \( S_N \), from \( n=5 \) to \( n=N \), is
\[
S_N = \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \ldots + \left( \frac{1}{N-1} - \frac{1}{N} \right)
\]
5. **Show Cancellation:**
Observe that this series telescopes such that most terms cancel:
\[
S_N = \frac{1}{4} - \frac{1}{N}
\]
6. **Calculate the Infinite Sum \( S \):**
As \( N \to \infty \), \( \frac{1}{N} \to 0 \), so the sum of the infinite series is:
\[
S = \frac{1}{4}
\]
**Solutions:**
- \( S_N = \frac{1}{4} - \frac{1}{N} \)
- \( S = \frac{1}{4} \)
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