8 6. Use the Root Test to determine whether Σ k=1 6k² 2k² +4 k converges absolutely or diverges.
8 6. Use the Root Test to determine whether Σ k=1 6k² 2k² +4 k converges absolutely or diverges.
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 16: Using the Root Test**
**Objective:**
Determine whether the infinite series \(\sum_{k=1}^{\infty} \left( \frac{-6k^2}{2k^2 + 4} \right)^k\) converges absolutely or diverges.
**Explanation:**
To solve this problem, we'll employ the Root Test, which is a method used to determine the convergence or divergence of an infinite series. The Root Test involves taking the k-th root of the absolute value of the k-th term of the series and then examining the limit of that sequence as \(k\) approaches infinity.
**Steps:**
1. Consider the general term of the series \(a_k = \left( \frac{-6k^2}{2k^2 + 4} \right)^k\).
2. Find the absolute value of the general term: \( |a_k| = \left( \left| \frac{-6k^2}{2k^2 + 4} \right| \right)^k = \left( \frac{6k^2}{2k^2 + 4} \right)^k\).
3. Apply the Root Test by considering the limit \( L = \lim_{k \to \infty} \sqrt[k]{|a_k|} \).
**Root Test:**
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \) or \( L = \infty \), the series diverges.
- If \( L = 1 \), the Root Test is inconclusive.
Next, we find the limit \( L \):
\[ L = \lim_{k \to \infty} \sqrt[k]{ \left( \frac{6k^2}{2k^2 + 4} \right)^k } \]
Breaking down:
\[ L = \lim_{k \to \infty} \left( \frac{6k^2}{2k^2 + 4} \right) \]
Simplify the fraction inside the limit:
\[ \frac{6k^2}{2k^2 + 4} = \frac{6k^2}{2k^2(1 + \frac{2}{k^2})} = \frac{6k^2}{2k^2 \cdot 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff7f42fd-3734-43a2-a946-a65fb76095a7%2Fbcca5eb8-9cd7-4996-b3c7-d761bda71fec%2Fn1n50no_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 16: Using the Root Test**
**Objective:**
Determine whether the infinite series \(\sum_{k=1}^{\infty} \left( \frac{-6k^2}{2k^2 + 4} \right)^k\) converges absolutely or diverges.
**Explanation:**
To solve this problem, we'll employ the Root Test, which is a method used to determine the convergence or divergence of an infinite series. The Root Test involves taking the k-th root of the absolute value of the k-th term of the series and then examining the limit of that sequence as \(k\) approaches infinity.
**Steps:**
1. Consider the general term of the series \(a_k = \left( \frac{-6k^2}{2k^2 + 4} \right)^k\).
2. Find the absolute value of the general term: \( |a_k| = \left( \left| \frac{-6k^2}{2k^2 + 4} \right| \right)^k = \left( \frac{6k^2}{2k^2 + 4} \right)^k\).
3. Apply the Root Test by considering the limit \( L = \lim_{k \to \infty} \sqrt[k]{|a_k|} \).
**Root Test:**
- If \( L < 1 \), the series converges absolutely.
- If \( L > 1 \) or \( L = \infty \), the series diverges.
- If \( L = 1 \), the Root Test is inconclusive.
Next, we find the limit \( L \):
\[ L = \lim_{k \to \infty} \sqrt[k]{ \left( \frac{6k^2}{2k^2 + 4} \right)^k } \]
Breaking down:
\[ L = \lim_{k \to \infty} \left( \frac{6k^2}{2k^2 + 4} \right) \]
Simplify the fraction inside the limit:
\[ \frac{6k^2}{2k^2 + 4} = \frac{6k^2}{2k^2(1 + \frac{2}{k^2})} = \frac{6k^2}{2k^2 \cdot 1
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