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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Can you help explain whether the series converges
![**Problem 29**
Express the infinite series as follows:
\[
\sum_{k=1}^{\infty} \frac{(-1)^k k}{6k^3 + 1}
\]
**Explanation:**
This series involves an alternating pattern due to the \((-1)^k\) term, which means the sign of each term alternates between positive and negative as \(k\) increases. The expression in the numerator is \(k\), while the denominator is a polynomial expression, \(6k^3 + 1\). The series starts from \(k = 1\) and continues to infinity. The behavior of such a series can be further analyzed for convergence using tests like the Alternating Series Test or the Ratio Test.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb8923159-503d-4bbc-a79d-55b83510ed02%2Faeed31d5-fcf8-48b0-9d94-1545736e5ba6%2Fjtapv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 29**
Express the infinite series as follows:
\[
\sum_{k=1}^{\infty} \frac{(-1)^k k}{6k^3 + 1}
\]
**Explanation:**
This series involves an alternating pattern due to the \((-1)^k\) term, which means the sign of each term alternates between positive and negative as \(k\) increases. The expression in the numerator is \(k\), while the denominator is a polynomial expression, \(6k^3 + 1\). The series starts from \(k = 1\) and continues to infinity. The behavior of such a series can be further analyzed for convergence using tests like the Alternating Series Test or the Ratio Test.
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