4. Find s, and the maximum error using the alternating series test for: (-1)*+1, (n + 1)! n=1
4. Find s, and the maximum error using the alternating series test for: (-1)*+1, (n + 1)! n=1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Problem Statement
4. **Find \( s_6 \) and the maximum error using the alternating series test for:**
\[
\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n^3}{(n+1)!}
\]
### Detailed Steps to Solve the Problem
#### 1. Understanding the Series
This series is an alternating series due to the \((-1)^{n+1}\) term.
#### 2. Alternating Series Test (Leibniz's Test)
The Alternating Series Test states that for a series of the form:
\[
\sum_{n=1}^{\infty} (-1)^{n+1} b_n
\]
where \( b_n \) is positive and decreasing, and \(\lim_{n \to \infty} b_n = 0\), the series converges. The error \( |s - s_n| \) when approximating the sum by \( s_n \) is less than or equal to the first omitted term, \( b_{n+1} \).
#### 3. Application to Given Series
Here, \( b_n = \frac{n^3}{(n+1)!} \).
Let's find \( s_6 \) and the error.
#### 4. Calculation of \( s_6 \)
The partial sum \( s_6 \) is the sum of the first 6 terms:
\[
s_6 = \sum_{n=1}^{6} (-1)^{n+1} \frac{n^3}{(n+1)!}
\]
Calculate each term individually:
- For \( n = 1 \): \( (-1)^2 \frac{1^3}{2!} = \frac{1}{2} \)
- For \( n = 2 \): \( (-1)^3 \frac{2^3}{3!} = -\frac{8}{6} = -\frac{4}{3} \)
- For \( n = 3 \): \( (-1)^4 \frac{3^3}{4!} = \frac{27}{24} = \frac{9}{8} \)
- For \( n = 4 \): \( (-1)^5 \frac{4^3}{5!} = -\frac{64}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4ebf9e4a-d88a-4068-a98e-7c5e269e8031%2F555288de-2c66-45fa-bebb-4a213294d7ab%2Fs641gi_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
4. **Find \( s_6 \) and the maximum error using the alternating series test for:**
\[
\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n^3}{(n+1)!}
\]
### Detailed Steps to Solve the Problem
#### 1. Understanding the Series
This series is an alternating series due to the \((-1)^{n+1}\) term.
#### 2. Alternating Series Test (Leibniz's Test)
The Alternating Series Test states that for a series of the form:
\[
\sum_{n=1}^{\infty} (-1)^{n+1} b_n
\]
where \( b_n \) is positive and decreasing, and \(\lim_{n \to \infty} b_n = 0\), the series converges. The error \( |s - s_n| \) when approximating the sum by \( s_n \) is less than or equal to the first omitted term, \( b_{n+1} \).
#### 3. Application to Given Series
Here, \( b_n = \frac{n^3}{(n+1)!} \).
Let's find \( s_6 \) and the error.
#### 4. Calculation of \( s_6 \)
The partial sum \( s_6 \) is the sum of the first 6 terms:
\[
s_6 = \sum_{n=1}^{6} (-1)^{n+1} \frac{n^3}{(n+1)!}
\]
Calculate each term individually:
- For \( n = 1 \): \( (-1)^2 \frac{1^3}{2!} = \frac{1}{2} \)
- For \( n = 2 \): \( (-1)^3 \frac{2^3}{3!} = -\frac{8}{6} = -\frac{4}{3} \)
- For \( n = 3 \): \( (-1)^4 \frac{3^3}{4!} = \frac{27}{24} = \frac{9}{8} \)
- For \( n = 4 \): \( (-1)^5 \frac{4^3}{5!} = -\frac{64}{
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