Problem 64
Question
Show that neither the Ratio Test nor the Root Test provides information about the convergence of $$\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{p}} \quad(p constant )$$
Step-by-Step Solution
Verified Answer
Neither the Ratio Test nor the Root Test can determine the convergence of the series; both are inconclusive.
1Step 1: Understand the Ratio Test
The Ratio Test states that for a series \( \sum a_n \), if \( L = \lim_{{n \to \infty}} \left| \frac{{a_{{n+1}}}}{{a_n}} \right| \), then the series converges if \( L < 1 \), diverges if \( L > 1 \), and the test is inconclusive if \( L = 1 \).
2Step 2: Apply the Ratio Test to the Given Series
Given the series \( \sum_{{n=2}}^\infty \frac{1}{(\ln n)^p} \), applying the Ratio Test involves calculating the following limit:\[ L = \lim_{{n \to \infty}} \left| \frac{{\frac{1}{(\ln(n+1))^p}}}{{\frac{1}{(\ln n)^p}}} \right|= \lim_{{n \to \infty}} \left( \frac{{(\ln n)^p}}{{(\ln(n+1))^p}} \right)\]Let's simplify this expression next.
3Step 3: Simplify the Ratio Limit
Simplifying the ratio gives:\[ L = \lim_{{n \to \infty}} \left( \left( \frac{{\ln n}}{{\ln(n+1)}} \right)^p \right)\]Since \( \ln(n+1) \sim \ln n \) as \( n \to \infty \), the limit becomes:\[ L = \lim_{{n \to \infty}} \left( 1 \right)^p = 1\]According to the Ratio Test, if \( L = 1 \), the test is inconclusive.
4Step 4: Understand the Root Test
The Root Test states for a series \( \sum a_n \), if \( L = \lim_{{n \to \infty}} \sqrt[n]{|a_n|} \), then the series converges if \( L < 1 \), diverges if \( L > 1 \), and the test is inconclusive if \( L = 1 \).
5Step 5: Apply the Root Test to the Series
For the series \( \sum_{{n=2}}^\infty \frac{1}{(\ln n)^p} \), calculate:\[ L = \lim_{{n \to \infty}} \sqrt[n]{\left| \frac{1}{(\ln n)^p} \right|}= \lim_{{n \to \infty}} \frac{1}{((\ln n)^p)^{1/n}}= \lim_{{n \to \infty}} \frac{1}{(\ln n)^{p/n}}\]As \( n \to \infty \), \( (\ln n)^{p/n} \to 1 \) because any fraction of \( \ln n \) raised to a power tending towards zero approaches one.
6Step 6: Conclude using the Tests
Therefore, for the Root Test, \( L = 1 \), which means the test is inconclusive. Both the Ratio Test and the Root Test provide no information about the convergence of the series.
Key Concepts
Ratio TestRoot TestLogarithmic Functions
Ratio Test
The Ratio Test is a powerful tool in determining the behavior of infinite series. It helps to decide whether a series converges or diverges based on the limit of the ratio of successive terms. For a series \( \sum a_n \):
- If \( L = \lim_{{n \to \infty}} \left| \frac{{a_{{n+1}}}}{{a_n}} \right| < 1 \), the series converges absolutely.
- If \( L > 1 \), the series diverges.
- If \( L = 1 \), the test is inconclusive, meaning the test provides no information about convergence.
Root Test
The Root Test focuses on the nth root of the absolute value of the series terms. It evaluates convergence similar to the Ratio Test but can sometimes be more effective. Given a series \( \sum a_n \):
- If \( L = \lim_{{n \to \infty}} \sqrt[n]{|a_n|} < 1 \), the series converges absolutely.
- If \( L > 1 \), the series diverges.
- If \( L = 1 \), the test is inconclusive.
Logarithmic Functions
Logarithmic functions are unique in their behavior at different scales. Understanding \( \ln n \) in the context of series can be tricky, as it grows very slowly compared to polynomial or exponential functions.
When dealing with series that involve logarithmic terms such as \( \sum_{{n=2}}^\infty \frac{1}{(\ln n)^p} \), it becomes important to capture how these terms interact with concepts of growth and decay.
When dealing with series that involve logarithmic terms such as \( \sum_{{n=2}}^\infty \frac{1}{(\ln n)^p} \), it becomes important to capture how these terms interact with concepts of growth and decay.
- \( \ln n \) approaches infinity as \( n \to \infty \), but at a very slow rate compared to \( n^p \) or \( e^n \).
- Series involving \( 1/(\ln n)^p \) can exhibit subtle convergence behavior, requiring deeper analysis beyond standard tests like Ratio and Root tests.
- Understanding the implications of logarithmic growth helps in applying the right tests or transformations.
Other exercises in this chapter
Problem 64
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