Determine Convergence of Series: Summation of 1 / (k^(4/9) * ln(k)) from k=3 to Infinity

Math Problem Statement

Determine whether the following series converges. Justify your answer. Summation from k equals 3 to infinity StartFraction 1 Over k Superscript four ninths Baseline ln k EndFraction Question content area bottom Part 1 Select the correct choice below and, if necessary, fill in the answer box to complete your choice. (Type an exact answer.) A. Let a Subscript kequalsStartFraction 1 Over k Superscript four ninths Baseline ln k EndFraction and b Subscript kequalsStartFraction 1 Over k ln k EndFraction . Since ModifyingBelow lim With k right arrow infinityStartStartFraction StartFraction 1 Over k Superscript four ninths Baseline ln k EndFraction OverOver StartFraction 1 Over k ln k EndFraction EndEndFraction equalsinfinity and Summation from k equals 3 to infinity StartFraction 1 Over k ln k EndFraction diverges, the series diverges by the Limit Comparison Test. B. Let a Subscript kequalsStartFraction 1 Over k Superscript four ninths Baseline ln k EndFraction and b Subscript kequalsStartFraction 1 Over k ln k EndFraction . Since ModifyingBelow lim With k right arrow infinityStartStartFraction StartFraction 1 Over k Superscript four ninths Baseline ln k EndFraction OverOver StartFraction 1 Over k ln k EndFraction EndEndFraction is nonnegative and Summation from k equals 3 to infinity StartFraction 1 Over k ln k EndFraction converges, the series converges by the Limit Comparison Test. C. The Ratio Test yields requals enter your response here. This is greater than 1, so the series diverges by the Ratio Test. D. The terms of the series are alternating and their limit is enter your response here, so the series diverges by the Alternating Series Test. E. The Ratio Test yields requals enter your response here. This is less than 1, so the series converges by the Ratio Test. F. The terms of the series are alternating and their limit is enter your response here, so the series converges by the Alternating Series Test.

Solution

To determine the convergence of the series

$\sum_{k=3}^{\infty} \frac{1}{k^{4/9} \ln k}$

we will analyze it using the Limit Comparison Test. The goal is to find an appropriate comparison series that has known convergence or divergence behavior.

Step 1: Choose a Comparison Series

Let’s define the series in terms of two sequences:

Let $a_k = \frac{1}{k^{4/9} \ln k}$ , which represents the terms of the original series.
Choose a comparison sequence $b_k = \frac{1}{k \ln k}$ .

We selected $b_k = \frac{1}{k \ln k}$ because it has a simpler exponent on $k$ (specifically, $k^1$ in the denominator) and we know the behavior of this series.

Step 2: Check Convergence or Divergence of $\sum b_k$

The series $\sum_{k=3}^{\infty} \frac{1}{k \ln k}$ is a known divergent series. This result comes from comparison with an integral, as the integral $\int \frac{1}{x \ln x} \, dx$ diverges.

Step 3: Apply the Limit Comparison Test

The Limit Comparison Test states that if

$\lim_{k \to \infty} \frac{a_k}{b_k} = L$

where $L$ is a positive, finite number, then $\sum a_k$ and $\sum b_k$ have the same convergence behavior.

Let’s compute the limit:

$\lim_{k \to \infty} \frac{a_k}{b_k} = \lim_{k \to \infty} \frac{\frac{1}{k^{4/9} \ln k}}{\frac{1}{k \ln k}} = \lim_{k \to \infty} \frac{k}{k^{4/9}} = \lim_{k \to \infty} k^{1 - 4/9} = \lim_{k \to \infty} k^{5/9} = \infty$

Since this limit is $\infty$ , the Limit Comparison Test indicates that the terms $a_k$ diverge faster than $b_k$ , which diverges. Therefore, the series $\sum a_k = \sum_{k=3}^{\infty} \frac{1}{k^{4/9} \ln k}$ also diverges.

Conclusion

The correct answer is: A. Since $\lim_{k \to \infty} \frac{a_k}{b_k} = \infty$ and $\sum_{k=3}^{\infty} \frac{1}{k \ln k}$ diverges, the series diverges by the Limit Comparison Test.

Do you want more details or have any questions about this solution?

Here are five related questions to expand your understanding:

How does the Limit Comparison Test differ from the Direct Comparison Test?
Why does the series $\sum_{k=3}^{\infty} \frac{1}{k \ln k}$ diverge?
How would we handle this problem if $k$ started from 1 instead of 3?
Could we use the Integral Test for this series? If so, how would it apply?
What other tests could be considered for this series, and why are they less effective?

Tip: For series with terms that decay polynomially and logarithmically, comparison tests with simpler terms are often efficient ways to analyze convergence behavior.

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Math Problem Analysis

Mathematical Concepts

Series Convergence
Limit Comparison Test
Divergent Series

Formulas

Limit Comparison Test formula: lim(k→∞) (a_k / b_k) = L

Theorems

Limit Comparison Test
Divergence of Harmonic Series with Logarithmic Terms

Suitable Grade Level

College-level Calculus

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