Chapter 9: Problem 43
Use any method to determine whether the series converges. $$\sum_{k=1}^{\infty} \frac{\sqrt{k} \ln k}{k^{3}+1}$$
Short Answer
Expert verified
The series converges by the Limit Comparison Test.
Step by step solution
01
Recognize the Form of the Series
The series given is \( \sum_{k=1}^{\infty} \frac{\sqrt{k} \ln k}{k^{3}+1} \). Our goal is to determine if this series converges or diverges. It's helpful to examine the expression for the general term as a function of \(k\), which will allow us to apply a suitable convergence test.
02
Simplify the Expression for Large k
For large \(k\), the term \(k^3 + 1\) behaves like \(k^3\). Thus, the general term can be approximated by \( \frac{\sqrt{k} \ln k}{k^3} = \frac{\ln k}{k^{2.5}} \). This simplification will help in applying the Limit Comparison Test.
03
Choose a Reference Series
Consider the series \(\sum_{k=1}^{\infty} \frac{\ln k}{k^{2.5}} \). This series is known to be similar to \(\sum_{k=1}^{\infty} \frac{1}{k^{2.5 - \epsilon}}\) for any small \(\epsilon > 0\), which converges because it is a \(p\)-series with \(p > 1\).
04
Apply the Limit Comparison Test
Use the Limit Comparison Test between the original series and the reference series \(\sum_{k=1}^{\infty} \frac{\ln k}{k^{2.5}}\). Calculate the limit: \[ \lim_{k \to \infty} \frac{\frac{\sqrt{k} \ln k}{k^3 + 1}}{\frac{\ln k}{k^{2.5}}} = \lim_{k \to \infty} \frac{\sqrt{k} k^{2.5}}{k^3 + 1} = \lim_{k \to \infty} \frac{k^3}{k^3 + 1} = 1 \]This limit is finite and positive, indicating that both series either converge or diverge together.
05
Conclusion about Convergence
Since \(\sum_{k=1}^{\infty} \frac{\ln k}{k^{2.5}}\) is a convergent series, and the Limit Comparison Test indicates that our original series behaves similarly, we conclude that the original series \(\sum_{k=1}^{\infty} \frac{\sqrt{k} \ln k}{k^{3}+1}\) also converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Comparison Test
The Limit Comparison Test is a valuable tool in determining the convergence or divergence of a series. This method involves comparing the series in question with a second, well-understood series. If we have two series, \( \sum a_k \) and \( \sum b_k \), where \( b_k \) is a reference series, the test requires us to find the limit of the ratio \( \frac{a_k}{b_k} \) as \( k \) approaches infinity.
To use the Limit Comparison Test:
To use the Limit Comparison Test:
- Choose a series \( \sum b_k \) that is similar to the original series \( \sum a_k \).
- Compute \( \lim_{k \to \infty} \frac{a_k}{b_k} \).
- If the limit is a positive, finite number, then both series will either converge or diverge together.
p-Series
The concept of a \( p \)-series is central in analyzing series convergence. A \( p \)-series is a series of the form \( \sum_{k=1}^{\infty} \frac{1}{k^p} \), where \( p \) is a positive constant. The convergence of a \( p \)-series depends on the value of \( p \):
- If \( p > 1 \), the series converges.
- If \( p \leq 1 \), the series diverges.
Convergence Tests
When dealing with infinite series, several convergence tests can be applied to determine whether a series converges or not. Each test is suited to a specific form of series and provides essential insights into its behavior.
Some common convergence tests include:
Some common convergence tests include:
- Comparison Test: Compare the series with another one whose convergence is known.
- Ratio Test: Use the ratio of successive terms to infer about convergence.
- Alternating Series Test: Applicable when series terms alternate in sign.
- Limit Comparison Test: Useful for comparing rates of growth between terms of two series.