Chapter 11: Problem 10
Which of the series in Exercises 1–36 converge, and which diverge? Give reasons for your answers. $$ \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{2}} $$
Short Answer
Expert verified
The series \( \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{2}} \) converges.
Step by step solution
01
Identify the Type of Series
The given series is \( \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{2}} \). We see that this is a series with terms involving natural logarithm, making it suitable for tests like the Integral Test.
02
Set Up the Integral Test
The Integral Test tells us that if \( f(n) = \frac{1}{(\ln n)^{2}} \) is a continuous, positive, decreasing function for \( n \geq 2 \), then \( \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{2}} \) converges if and only if the integral \( \int_{2}^{\infty} \frac{1}{(\ln x)^{2}} \, dx \) converges.
03
Check Conditions for the Integral Test
The function \( f(n) = \frac{1}{(\ln n)^{2}} \) is continuous, positive, and decreasing for \( n \geq 2 \) because \( \ln n \) becomes larger as \( n \) increases, making \( (\ln n)^2 \) larger and \( \frac{1}{(\ln n)^{2}} \) smaller.
04
Evaluate the Integral
Compute the integral \( \int_{2}^{\infty} \frac{1}{(\ln x)^{2}} \, dx \). Let \( u = \ln x \), then \( du = \frac{1}{x} \, dx \), which implies \( dx = x \, du = e^u \, du \). Thus, the integral becomes \( \int_{\ln 2}^{\infty} \frac{1}{u^2} \cdot e^{u} \, du \). This simplifies to \( \int \frac{1}{u^2} \, du = \frac{-1}{u} \). Evaluate from \( u = \ln 2 \) to \( u \to \infty \).
05
Evaluate the Limits
Evaluating \( \left[ \frac{-1}{u} \right]_{\ln 2}^{\infty} \) gives \( \lim_{u \to \infty} \left( \frac{-1}{u} \right) - \left( \frac{-1}{\ln 2} \right) = 0 + \frac{1}{\ln 2} \). Since the limiting process as \( u \to \infty \) indeed results in a finite value, the integral is convergent.
06
Conclusion Based on Integral Test
Since the integral \( \int_{2}^{\infty} \frac{1}{(\ln x)^{2}} \, dx \) converges, by the Integral Test, the series \( \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{2}} \) also converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Test
The Integral Test is a helpful tool for determining the convergence of a series. It's especially useful for series where the terms diminish in size in a specific manner, such as those involving logarithmic functions.To apply the Integral Test, you need to ensure the function you're dealing with, say \( f(n) \), meets these criteria:
- It's continuous, meaning no sudden jumps or breaks.
- It's positive for the interval you're considering, which makes things more straightforward.
- It's decreasing, which tells us the terms are becoming progressively smaller.
Natural Logarithm Function
The natural logarithm function, written as \( \ln(x) \), is a crucial part of mathematical analysis. It's the inverse of the exponential function \( e^x \), where \( e \) is approximately 2.71828. The relation \( y = \ln(x) \) means that \( e^y = x \).Key properties of the natural logarithm function are:
- \( \ln(x) \) is defined for all positive \( x \).
- It's an increasing function, meaning \( \ln(x) \) continues to grow as \( x \) gets larger.
- The logarithm of 1 is 0, i.e., \( \ln(1) = 0 \).
Convergent Series
A convergent series is one where the sum of its terms approaches a specific finite value as you add more and more terms. It's a central concept in calculus and mathematical analysis.In the case of the series \( \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{2}} \), we determined its convergence using the Integral Test. By substituting the series with the integral \( \int_{2}^{\infty} \frac{1}{(\ln x)^{2}} \, dx \), we check if the integral achieves a finite result.Steps to determine convergent series:
- Ensure that the integral or test applied confirms a finite sum.
- Once established, conclude the original series also reaches a finite sum.